Download - Te Kura

te mĀtauranga pĀngarau te tauanga
mathematics and statistics
MX2062
algebraic methods 2
ncea level 2
2012/1
mathematics and statistics
ncea level 2
Expected time to complete work
This work will take you about 10 hours to complete.
You will work towards the following standard:
Achievement Standard 91261 (Version 1) Mathematics and Statistics 2.6
Apply algebraic methods in solving problems
Level 2, External
4 credits
Achievement Standard 91269 (Version 1) Mathematics and Statistics 2.14
Apply systems of equations in solving problems
Level 2, Internal
2 credits
In this booklet you will focus on these learning outcomes:
• manipulate algebraic expressions
• determine the nature of the roots of a quadratic equation
• forming, use and solve linear and quadratic equations
• solve simultaneous linear and non-linear equations.
You will continue to work towards Achievement Standard 91261 (2.6) in booklet MX2063.
You will continue to work towards Achievement Standard 91261 (2.14) in booklet
MX2063. The other work is covered in MX2021, MX2022 and MX2023.
Copyright © 2012 Board of Trustees of Te Aho o Te Kura Pounamu, Private Bag 39992, Wellington Mail Centre, Lower Hutt 5045,
New Zealand. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means without the
written permission of Te Aho o Te Kura Pounamu.
© te ah o o te k u ra p ou n am u
contents
1
Equation-solving techniques
2
Translating words into mathematics
3
Quadratic equations
4
Problems involving quadratic equations
5
Problems involving simultaneous equations
6
Review activity
7
Answer guide
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how to do the work
When you see:
1A
Complete the activity.
Check your answers.
Your teacher will assess this work.
Use a calculator.
Contact your mathematics and statistics teacher.
You will need:
•• paper to work on, an exercise book with squared paper is recommended
•• a scientific or graphical calculator
•• access to OTLE, but this is not essential.
Resource overview
You will need to write the answers on your own paper.
A graphics calculator is recommended but is not essential. Using a graphics calculator
will enhance your understanding and is permitted in the assessment activities.
At the end of each exercise, mark your work using the Answer guide. The answers will give
you useful feedback and help you in the learning process.
You should telephone or email your mathematics and statistics teacher if you would like to
discuss your work.
At the completion of this booklet and your work on the review activity, complete the
self-assessment page and the cover sheet, and send these to your mathematics and statistics
teacher to assess. Your self-assessment will be returned to assist you with revising for
these achievement standards.
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1
equation-solving techniques
learning outcomes
Form, use and solve linear and quadratic equations.
Solve simultaneous linear and non-linear equations.
learning intentions
In this lesson you will learn to:
•• manipulate algebraic equations to solve
•• form, use and solve linear and quadratic equations
•• solve simultaneous equations by substitution, graphing and elimination methods.
introduction
This booklet is about using the concepts of algebra you have learnt in the first booklet and
developing these ideas to solve problems.
You will learn how to translate words into mathematical symbols and how to set up
equations to model situations.
You will use methods for solving linear and quadratic equations, and review how to solve
simultaneous equations both linear and quadratic.
revision: linear and quadratic equations
Solve for x:
1A
1.2(x – 5) = 5x + 2
2.3(3x + 2) – 2(x – 3) = 5
3.
x
x
=
–6
8
2
4. x² – 5x + 6 = 0
5. x² = 3x
6.3x² = 7x + 6
7. x + 12 = x²
Check your answers.
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equation-solving techniques
simultaneous equations
All of the equations you have solved so far have had only one variable. Simultaneous equations
involve two variables. They are called simultaneous equations because a solution is found that
satisfies two (or more) equations at the same time.
The simplest form of simultaneous equation is two linear equations;
for example, y = 2x – 5 and 2x + 3y = 9. We shall deal with this type first.
There are three main methods of finding a solution:
• graphing
• elimination
• substitution (we deal with this next lesson).
graphing
The method here is to draw the graph of both equations on the same pair of axes and then read
the x and y coordinates off the point of intersection.
Example 1
Find the solution of the equations y = 2x – 5 and 2x + 3y = 9.
The equation y = 2x – 5 is written in the form y = mx + c, so we can say directly that
this line has:
gradient (m) = 2 and y-intercept (c) = –5
The equation 2x + 3y = 9 is more suited to the two-intercept method, because we can use x = 0
and y = 0 to find where the line cuts each axis.
If x = 0, 2 × 0 + 3y = 9
y
y=3
y = 2x – 5
3
If y = 0, 1
x=4
2
2x + 3 × 0 = 9
Now draw both lines on one pair of axes
and read off the point of intersection.
The lines meet at (3, 1), so the solution is
x = 3, y = 1.
4
x
0
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-
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equation-solving techniques
1B
Solve these simultaneous equations using the graphing method.
1. y = 3x – 32.
y = 2x + 6
y = –2x + 72y + x = 7
Check your answers.
elimination
The second method for solving simultaneous equations is called elimination.
In this method, you first multiply each equation by a number that will give the coefficient
of either the x terms or the y terms the same value but opposite signs. Then you add the two
equations. This eliminates either the x terms or the y terms and makes it possible to find a
value for the remaining variable.
Before starting this process, check that you can do all the necessary steps individually.
Answer these questions then check your answers with those that follow.
revision
1. What is the coefficient of y in 3x – 2y = 5?
2. Multiply both sides of these equations by 4.
a. 3x – 2y = 5
b. x + 4y = 2
c. –2x – y = 1
3. Multiply both sides of these equations by –2.
a. 2x + 5y = –11
b. –3x – 2y = 4
c. x + 4y = 7
Solutions:
1.–2
2. a. 12x – 8y = 20
b. 4x + 16y = 8
c. –8x – 4y = 4
3. a. –4x – 10y = 22
b. 6x + 4y = –8
c. –2x – 8y = –14
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equation-solving techniques
Example 2
Solve by elimination:
x + 2y = 6
x – y = 3
(A)
(B)
We will use a letter to represent each equation for reference.
We could eliminate x or y, but this time we’ll eliminate y.
To eliminate y, these coefficients need to have the same absolute value, but the opposite sign.
Multiplying equation (B) by 2 makes the coefficient of its y term equal –2. This has the same
absolute value but the opposite sign to the coefficient of y in equation (A), as required.
Multiply equation (B) by 2: 2x – 2y = 6
Rewrite equation (A)
x + 2y = 6
(C)
(A)
Next, add equations (C) and (A) to eliminate y.
2x + x – 2y + 2y
3x + 0y
3x x = 6 + 6
= 12
= 12
=4
Notice how the y terms were eliminated by adding.
Now find y by substituting 4 for x in equation (A) or (B).
In (A): x + 2y = 6
4 + 2y = 6
4 + 2y – 4 = 6 – 4
2y = 2
y = 1
The solution is (4, 1).
As you found y by substituting in equation (A), you should do your check from equation (B).
x – y = 4 – 1
= 3 
This pair of equations could also be solved by multiplying equation (B) by –1 to produce
coefficients for x of +1 and –1. You would then add the equations, eliminating x, and solve the
resulting linear equation for y.
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equation-solving techniques
Example 3
We will use the same equations in example 2 but this time we will eliminate x.
x + 2y = 6
(A)
x + y = 3
(B)
Multiply (B) by –1: –x + y = –3 (C)
Rewrite equation (A)
x + 2y = 6
Add (C) and (A) to eliminate x.
(A)
3y = 12
y = 1
Now find x by substituting y = 1 into (A) or (B).
x + 2(1) = 6
x+2 = 6
x = 4
The solution is (4, 1).
Check in (B) x – y = 4 – 1
=
3 ✓
Example 4
Find the solution of these simultaneous equations:
3x – 2y + 4 = 0
(A)
2x + 5y + 28 = 0
(B)
This time we shall eliminate x.
Multiply (A) by 2:
Multiply (B) by 3:
–
6x – 4y + 8 = 0
(C)
6x – 15y – 84 = 0
(D)
–
Add to eliminate x:
19y – 76 = 0
–
–
Substitute y = –4 into (B):
y = –4
2x + 5 × –4 + 28
= 0
2x – 20 + 28 = 0
2x + 8 = 0
19y = 76
2x = –8
2x ÷ 2 = –8 ÷ 2
x = –4
The solution is (–4, –4).
Check in (A) 3x – 2y + 4 = 3 × –4 – 2 × –4 + 4
–
= 12 + 8 + 4
= 0✓
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equation-solving techniques
1C
1. Solve this pair of simultaneous equations by elimination.
4x – 3y = 7
2x + 3y = –1
2. Solve by elimination.
a.7x – 2y = 3b.
2x + y = 10
–
3x + 8y = 433x + 5y = 29
c.2x – 3y = 1d.
3x + 2y = 13
3x – 4y = 77x + 3y = 27
Check your answers.
substitution
The aim of the substitution method is to develop one equation with only one variable.
This can be x or y.
First, ensure that one of the equations is written in the form y = …
Then substitute or replace the variable y in the other equation with the expression that y equals.
The resulting linear equation in x can then be solved. Finally, the value of y is found.
Example 5
Find the solution of the simultaneous equations y =3x – 2
y=x + 1
y = 3x – 2 (A)
Number the equations for reference.
y = x + 1 (B)
x + 1 = 3x – 2
The y values are the same in each equation,
so the y in equation (A) can be replaced
with x + 1 from equation (B).
1 + 2 = 3x – x Collect like terms.
3 =
2x 3
x = 2
Solve for x.
Find the value of y by substituting the x value
into either of the original equations.
From (B): y = x + 1
3
soy = 2
5
y=
2
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equation-solving techniques
It is a good idea to check your solution by substituting both values for x and y into the
other original equation.
y = 3x – 2
3
= 3 ×
– 2
2
5
= as required.
2
3 5
3
5
The solution is x = , y =
, which we may also write as ( , ).
2 2
2
2
In (A), Sometimes you may need to rearrange one of the equations before you can substitute.
Example 6
Solve the following pair of simultaneous equations.
x + 3y = 11
3x + 2y = 12
(A)
(B)
Start by rearranging the first equation to obtain an expression for x.
x + 3y = 11 gives x = 11 – 3y
Continue by substituting 11 – 3y for x in equation (B).
3x + 2y = 12
3(11 – 3y) + 2y = 12
Now solve for y:
33 – 9y + 2y
33 – 7y
33 – 7y – 33
–7y
y
(B) becomes
= 12
= 12
= 12 – 33
= –21
= 3
Now substitute to find x:
In (A):
x + 3y = 11
x + 9 = 11
x = 2
Check your answers in equation (B):
3x + 2y = 12
3 × 2 + 2 × 3 = 12
6 + 6 = 12
12 = 12
The solution of the pair of equations is (2, 3).
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equation-solving techniques
1D
Solve the following pairs of simultaneous equations.
1.a. y = 4x + 1
y = x + 4
(A)
(B)
Hint: Substitute x + 4 for y in (A).
b.y = 2x – 3
y = 4x + 5
2. a. 2x + 3y = 1
y = 2x – 5
(A)
(B)
Hint: Substitute 2x – 5 for y in (A).
b. 2x – 3y = 11
y = 3x – 6
(A)
(B)
c. 3x + 4y = 31
2x = y + 6
(A)
(B)
Hint: Rewrite equation (B) with y as the subject.
d.
6x – 3y = –8
x + 2y = 7
Check your answers.
Which method to use?
The substitution method is generally the most useful. It can be applied regardless
of the way the equations are written.
The elimination method is useful where neither of the coefficients of the x or y terms is 1.
The graphical method gives a very good visual of the linear equations and where the intersection point is.
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translating words into mathematics
learning outcome
Form, use and solve linear and quadratic equations.
learning intentions
In this lesson you will learn to:
•• change word problems into mathematical expressions or equations.
•• form and solve linear equations.
introduction
If you were translating from English into a foreign language, you’d probably start with the
words you know, then you’d put phrases together, and finally sentences.
EnglishSpanish
WordsThe catEl gato
PhrasesOn the matEn la estera
Sentences The cat is sitting El gato está sentado
on the maten la estera
In a similar way, words, phrases and sentences can be translated into the language of
mathematics.
EnglishMathematics
WordsFour4
PhrasesThree plus four3 + 4
Sentences The sum of three 3 + 4 = 7
and four is seven
In this lesson, you will practise translating from English to Mathematics. We’ve assumed you
know most of the words, but there are a couple of special cases where quite lengthy English
phrases can be translated into a single mathematical term.
EnglishMathematics
I’m thinking of a numberx
A certain numberx
The variablex
Here are some examples of phrases.
EnglishMathematics
The sum of a and ba + b
The product of seven and a7a
Twice as big as x2x
1
y
One third of y y or
3
3
g is increased by four g + 4
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2A
1. Start with the number x, and write down the results of each of the following operations.
a. Multiply it by two
b. Multiply it by minus one half
c. Divide it by four
d. Add three to it
e. Subtract five from it
f. Subtract it from five
g. Multiply it by two and then add three to the result
h. Add three to it and then multiply the result by two
i. Subtract five and then divide the result by two
j. Divide by seven and then add three to the result
2. Write these phrases using mathematical symbols.
a. seven more than x
b. The product of b and seven
c. six decreased by m
d.
g multiplied by two
e.Double x and add two
f. Half of a
g. Three more than twice h
h.
x decreased by four
i. One third of y
j. Five less than twice p
hx = xh
Check your answers.
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translating words into mathematics
translating sentences
As you do this section, concentrate on the translations. Do not solve the equations.
Cover the right-hand column of the table with a piece of paper. See if you can write each
word sentence using mathematical symbols. Check each answer before attempting the next.
We’ve used a for the variable in number 1, b in number 2, and so on.
Word sentence
Mathematics
1. A number multiplied by five equals 10. 5a = 10
or
etc
2. Three more than a number is equal to 17. b + 3 = 17or x + 3 = 17 etc
3. I think of a number and subtract three; the result
is the same as multiplying the original number by four. c – 3 = 4corp – 3 = 4p etc
4. I think of a number and square it; the result is 20
more than the original number.
d² = d + 20 or
2B
5x = 10 x2 = x + 20 etc
1. Write each of the following sentences as a mathematical equation.
Do not solve the equation.
a. If two is taken from z, the result is four.
b. The sum of q and six is 10.
c. If four is added to m, the result is nine.
d. The product of n and seven is 21.
e.If x is taken from 17, the result is four.
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translating words into mathematics
2. Write each statement as an equation involving x.
Do not solve the equation.
a. I think of a number, multiply it by three and add seven; the result is 42.
b. I think of a number, subtract five and then square the result. The final answer is 36.
c. I think of a number and add six; the result of doing that is the same as multiplying the original number by three.
d. I think of a number and add it to its reciprocal. The result is 4.25.
e. I think of a number, square it and then subtract it from twice the original number.
The result is –32.
3. For each of the following:
i. Introduce and define two variables.
ii. Form two equations from the conditions in the problem.
Do not solve the equation.
a. I’m thinking of two numbers. Their sum is 42 and their difference is three.
b. I’m thinking of two numbers. Their sum is 27 and their product is 50.
c. I’m thinking of two numbers. Their sum is four and the sum of their reciprocals is –0.3.
Check your answers.
Each of the translations you have done has been based on mathematical statements, where the
words talk about numbers. The next few statements are more practical; they come from real-life
situations. In each case, you will have to decide what you are asked to find, introduce a letter
or variable for this, and state clearly what it represents. Again, concentrate on the translation;
solving these equations will come later.
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translating words into mathematics
Example 1
Write a mathematical sentence for the following problem.
In three yearsʼ time, Bill will be 46. How old is he now?
Solution
You are asked to find out how old Bill is now, so start by introducing a letter for this.
Write:Let x years be Bill’s age now.
Now translate the words into mathematical symbols:
x + 3 = 46
By solving this equation, you could find Bill’s age, but that is not important at this stage.
Concentrate on the process leading to finding the equation.
Example 2
Translate into mathematics:
There were 18 dinner mints in a box. They were passed around at a dinner party and each
person had two. There were four left. How many people were at the dinner party?
Solution
Introduce a letter Let the number of people be p.
for the unknown quantity.
Think.If one person eats 2 dinner mints, then p people eat 2p dinner mints.
Translate.2p + 4 = 18
The mints eaten the mints left the number in the box.
Solving this equation would tell you how many people were at the dinner party.
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translating words into mathematics
2C
1. In the van Heusen family, Sonja is m years old. Complete each of the following statements about the ages of the other members of the family with mathematical symbols.
a. Pieter is twice as old as Sonja, so Pieter’s age is
.
b. Hans is five years younger than Sonja, so Hans’s age is
c. Wilhemina is three years older than Pieter, so Wilhemina’s age is d. Marita is two years younger than Hans, so Marita’s age is
.
.
.
2. For each of the following problems:
i. Introduce a letter and state clearly what it represents.
a.
b.
c.
d.
e.
f.
ii. Write an equation (but do not solve it).
The distance from Upper Hutt to Wellington is 34 km. Seth has driven 15 km. How far has he still to go?
In a test, the top mark was four times as high as the bottom mark.
If the top mark was 48, what was the bottom mark?
There were 24 pieces in a block of chocolate. The children each had three pieces
and there were three pieces left over. How many children shared the chocolate?
Joan had a quick way of remembering the postal code for her Australian pen friend.
It is five more than five times 500. What is the postal code?
The length of a rectangular lawn is 18 m.
If the perimeter of the lawn is 56 m, how wide is it?
At a sheep sale, George sold 50 more sheep than Charlie.
Together they sold 160 sheep. How many sheep did Charlie sell?
Check your answers.
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problems involving linear equations
Some people have an intuitive approach to problem solving; others work by trial and error.
However, a systematic, step-by-step approach is usually much more efficient and much more
likely to lead to a successful solution.
We recommend that you use these steps in all of the following problems, even if you can see
the answer straight away. By practising the routine on simple problems, you will develop the
confidence to use it on more difficult problems.
The steps are:
1. Read through the problem carefully. Try restating it in your own words.
2. Consider drawing a diagram.
3. Choose a letter for the unknown number or quantity and state clearly what
it represents.
4. Form an equation from the conditions in the problem.
5. Solve the equation.
6. Check your answer with the conditions stated in the problem.
7. Write an English sentence answering the original question.
Example 3
A rectangle is three times as long as it is wide.
Its perimeter is 32 cm. What is its width?
Solution
1. Read through the problem carefully.
Think: The length is three times the width, and the distance right around is 32 cm.
2. Consider drawing a diagram.
Can you draw one that illustrates this problem?
Yes. This will be useful in step 4.
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3. Choose a letter for the unknown number and state clearly what it represents.
It’s usually a good idea to let the letter stand for the number or quantity you are trying to find. In this case you’d write:
Let the width of the rectangle be x cm.
4. Form an equation from the conditions in the problem.
This is where your diagram can help.
The width is x.
3x
x
The length is ‘three times as
long as it is wideʼ.
x
3x
length = 3x
The perimeter is 32 cm. To find the perimeter, add the lengths of the four sides.
The equation is x + 3x + x + 3x = 32.
5. Solve the equation.
x + 3x + x + 3x = 32
8x = 32
x = 4
6. Check your answer with the conditions stated in the problem.
The width is 4 cm
The length is 3 × 4
=
12 cm
The perimeter is 4 + 12 + 4 + 12
=
32 cm
7. Write a sentence answering the original question.
The width of the rectangle is 4 cm.
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Example 4
Keri paid $7.75 when she bought a roll, some fruit and a can of juice for lunch.
She paid $3.25 more for the fruit than the roll and $1.75 more for the roll than the juice.
How much did the roll cost?
You may like to try solving this problem yourself using the guidelines shown in Example 3.
If not, work through the following questions.
This will help you learn the problem-solving process.
1. What are you asked to find?
2. Choose a letter for the unknown number and state clearly what it represents.
3. Write a mathematical statement for the price of the fruit.
4. Write a mathematical statement for the price of the juice.
5. Write an equation that connects the separate prices of the roll, fruit and juice,
and the total price.
6. Solve your equation in (5).
7. a. What is the price of the roll?
b. What is the price of the fruit?
c. What is the price of the juice?
d. What is the total cost of the lunch?
e. Do these answers agree with the information in the question?
8. Write a sentence answering the original problem.
Answer to example 4
1. The price of the roll.
2. Let the price of the roll be r $.
3.
Price of fruit = r – 3.25
4.
Price of juice = r – 1.75
5. r + (r – 3.25) + (r – 1.75) = 7.75
6. 3r – 5 = 7.75
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3r = 12.75
r = 4.25
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7. a. The roll costs $4.25
b. The fruit costs $1.00
c. The juice costs $2.50
d. The total cost is $7.75
e.Yes.
8. The cost of the roll is $4.25
In practice, you’d write solutions like this.
Example 5
The sum of four consecutive integers is 3 810. Find the integers.
Consecutive means they come one after the other.
5, 6, 7 and 8 are four consecutive integers.
Let p represent the first integer.
p + (p + 1) + (p + 2) + (p + 3) = 3 810
p + p + 1 + p + 2 + p + 3 = 3 810
4p + 6 = 3 810
4p = 3 804
p = 951
So p + 1 = 952
p + 2 = 953
p + 3 = 954
Check 951 + 952 + 953 + 954 = 3810
The four consecutive integers that add to 3810 are 951, 952, 953 and 954.
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2D
In the following problems, all variable(s) must be defined appropriately.
1. A rectangular lawn is twice as long as it is wide. Its perimeter is 96 m.
What is its width?
2. The sum of five consecutive whole numbers is 725. What is the first number?
3. Ken paid $8.80 when he bought a pie, a cake and a bottle of cola for lunch.
If the cake cost 40 c more than the cola, and the pie cost twice as much as the cake, how much was the cake?
4. A motorist travels at a certain speed for two hours, and then, after being fined for speeding,
does the next three hours at 15 k/m less.
If the total distance travelled is 480 kilometres, what was his speed for the first two hours?
If you have trouble doing this question by yourself, work through the following steps:
1. Choose a letter for the unknown value and state what it represents.
2. Write a mathematical statement for the distance travelled in the first two hours.
(Hint: Distance = speed × time)
3. Write a mathematical statement for the speed at which the motorist was travelling
after he got his ticket.
4. Write a statement for the distance travelled in the last three hours.
5. Write an equation for the total distance travelled.
6. Solve the equation.
7. Write a sentence answering the original question.
8. Check your answer.
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5. One tree is seven times as old as another. Five years from now, it will be four times
as old as the younger tree is then. How old is the younger tree now?
Hint:
a. If the younger tree is x years old now:
i. how old is the older tree now?
ii. how old is the younger tree in five yearsʼ time?
iii.how old is the older tree in five yearsʼ time?
b. Write an equation connecting the ages of the trees in five yearsʼ time.
c. Solve the equation. Check your answer and write a final sentence.
6. Supplementary angles add to 180°. If one of two supplementary angles is 3º smaller
than twice the other, what is the size of the smaller angle?
7. Several years ago Jeremy Jones joined a club. The entrance fee was $35.
The yearly subscription for the first three years was $240 and after that it rose to $336. He calculated that he’d spent a total of $2 099. How many years ago did he join the club?
Check your answers.
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3
quadratic equations
learning outcomes
Form, use and solve linear and quadratic equations.
Determine the nature of the roots of a quadratic equation.
learning intentions
In this lesson you will learn to:
•• form and solve quadratic equations using the quadratic formula
•• determine the nature of the roots of a quadratic equation using the discriminant.
introduction
Many problems involve other types of equations. In this lesson, you’ll look at quadratic
equations and how to solve them using the quadratic formula.
2x² + x – 3 = 0 and x² + 5x + 6 = 0 are examples of quadratic equations.
The method you have probably used to solve quadratic equations so far is based on
factorising (using brackets).
Example 1
Solve x² + 5x + 6 = 0
(x + 3)(x + 2) = 0
Either x + 3
or
x + 2 = 0
x = 3 or
The solutions are x = −3 and x = −2.
x=–2
= 0
–
Example 2
Solve 2x² + x – 3 = 0
2x² + x – 3 = 0
(2x + 3)(x – 1) = 0
or
x – 1 = 0
3 or
3
x = –
2
3
The solutions are x = −
and x = 1.
2
x =1
Either2x + 3
= 0
2x
=
–
In real life problems, many quadratic equations cannot be factorised.
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quadratic equations
A useful formula helps deal with these cases.
For any quadratic equation in the form ax² + bx + c
then
b±
−
x
=
= 0
b² − 4ac
2a
The symbol ± is read as ‘plus or minusʼ. You can write the two solutions as:
−
b + b² − 4ac
x = 2a
or x =
b−
−
b² − 4ac
2a
You can use this formula for any quadratic equation, but for those that are easy to factorise,
you may find factorising a quicker method.
If you’d like to see how this formula is formed, look in the appendix that is included
after the Answer guide.
remember!
ax² + bx + c = 0
−
b±
then x = If b² − 4ac
2a
Example 3
Solve 2x² + 6x + 3 = 0
ax² + bx + c = 0
a = 2, b = 6, c = 3
Compare the equation
with ax2 + bx + c = 0
to find a, b and c.
−
b ± b² − 4ac
x = 2a
x =
x =
Write the formula.
6 ± 6² − 4 × 2 × 3
2×2
−
Subtitute the values for a, b and c.
6 ± 36 − 24
4
−
−
6 ± 12
x =
4
The two solutions are:
−
6 + 12
x =
4
x = −0.6339746
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or
x =
or x =
6−
12
−
4
2.3660254
−
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quadratic equations
Round your solutions sensibly to put them in the context of the problem.
x = –0.63 or x =
–
2.37 (2 d.p.)
Follow the next examples through before attemping activity 3A.
Use your calculator to find decimal approximations for the solutions.
Example 4
Solve x² – 3x + 1 = 0
a = 1, b = –3, c = 1Write out the values for a, b and c.
−
b ± b² − 4ac
x = Write the formula.
2a
− −
( 3) ± (−3)² − 4 × 1 × 1
x = Substitute in a, b and c.
2×1
3± 5
x = Simplify.
2
x =
3+ 5
2
or
x =
3−
5
2
x = 2.62 or x = 0.38 (2 d.p.)
Any variable can be used instead of x. Just change x in the formula to a different variable.
Example 5
Solve 2p² + 5p = 2
2p² + 5p – 2 = 0
Move all terms to the left of the = sign.
a = 2, b = 5, c = –2Write out the values for a, b and c.
−
b ± b² − 4ac
p = Write the formula.
2a
p =
−
5 ± 5² − 4 × 2 × −2
Substitute a, b and c.
2×2
−
5 ± 25 + 16
p = Simplify.
4
p =
5±
41
−
4
−
5 + 41
4
or
p=
p = 0.35 or p =
p =
5−
41
−
4
–
2.85 (2 d.p.)
Now try the following exercise to practise using the quadratic formula.
Note: The solutions of quadratic equations are frequently called roots.
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quadratic equations
3A
Solve these equations using the formula.
Remember to check that all terms are on the left-hand side of the equals sign first.
If not, rearrange the equation, setting the right-hand side to zero, as in the last example.
1. x² + 5x + 3 = 0
2. 4x² + 9x – 4 = 0
3. x² – 3x – 2 = 0
4. 7s² – 3s = 1
5. x² – 6x + 5 = 0
6. 4x² = –12x – 9
Check your answers.
the nature of the roots of a quadratic equation
The solutions to quadratic equations are sometimes called roots.
For example, the roots of (x – 3)(x + 2) = 0 are 3 and –2.
As you will see in this booklet, it is often useful to know what the roots of a quadratic equation
are like without having to solve the equation. This is called finding or describing the nature of
the roots.
If you solved lots and lots of quadratic equations, you would find that the equations could be
divided into four different types, grouped according to the nature of their roots.
1. Equations with two irrational roots; for example,
7s² – 3s=1
3 ± 37
s =
14
37 is a positive number, so it has a square root.
There are two real roots:
3 + 37
s = 14
But
and s =
3 − 37
14
37 cannot be found exactly.
37 is a surd, and so the two roots are irrational.
2. Equations with two rational roots; for example,
x² – 6x + 5 = 0
6 ± 16
x =
2
16 is a positive number, so it has a square root. There are two real roots:
6 + 16
2
x =
Now
and x =
So the two roots are rational.
26
16
2
16 = 4, so the roots can be found exactly:
6+4
x = 2
x = 5
6−
6−4
2
and x = 1
and x =
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quadratic equations
3. Equations with two equal rational roots; for example,
4x² + 12x + 9 = 0
-12 ± 0
x =
8
0 = 0, so there is only one value for x.
x =
12
8
−
–
=
1.5
So there are two equal rational roots.
Alternatively, you can think of this equation as having one repeated root or two equal roots.
4. Equations with no real roots; for example,
3y²=y – 1
1 ± −11
y =
6
11 is less than zero, so it has no real square root. y cannot be calculated in this case.
There are no real roots.
–
the discriminant
The part of the formula that determines the nature of the roots is the expression under the square
root sign, b² – 4ac. Because b² – 4ac distinguishes, or discriminates, between different kinds
of roots, it is called the discriminant of the quadratic equation ax² + bx + c = 0.
the discriminant of a quadratic equation
For the equation ax² + bx + c = 0
the quadratic formula is x =
b±
b² − 4ac
2a
The discriminant b² – 4ac is the part of the formula under the
−
sign.
You will use the discriminant to find the number of real solutions of a quadratic
equation ax² + bx + c = 0.
b² – 4ac > 0
When the number under the
two distinct real solutions:
sign in the quadratic formula is positive, there will be
−
b + b² − 4ac
b − b² − 4ac
or x =
2a
2a
b² – 4ac = 0
−
b
When the square root part of the formula is zero, there will be two real and equal solutions: x = 2a
b² – 4ac < 0
Here, the number under the
sign in the formula is negative. It is not possible to find
the square root of a negative number, so there are no real solutions.
x =
−
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quadratic equations
Example 6
Find the nature of the roots of x² + 7x + 10 = 0.
Solution
First, compare with the general equation to find a, b and c.
x² + 7x + 10= 0
ax² + bx + c = 0
a = 1, b = 7, c = 10
Now find the value of b² – 4ac.
b² – 4ac = 7² – 4 × 1 × 10
= 49 – 40
= 9
b² – 4ac > 0, so there are two roots.
Nine is the square of a rational number, so the roots are rational. Therefore, x² + 7x + 10 = 0
has two rational roots.
Example 7
Find the value of the discriminant of 6x² + 2x + 5 = 0 and state the number of real solutions
for the equation.
a = 6,
b = 2,
c=5
The discriminant b² – 4ac = 2² – 4 × 6 × 5
–
=
116
b² – 4ac < 0, so there are no real solutions.
3B
1. The discriminant of 9x² – 2x + 1 = 0 is –32.
What is the nature of the roots?
2. The value of b² – 4ac for a particular quadratic equation is 0. Describe the roots.
3 The discriminant of 2x² – 8x – 1 = 0 is 72. What information does this give about
the nature of the roots?
4. Find the nature of the roots of each of the following equations.
a.
x² + 3x – 10 = 0
b.
x² + 3x – 11 = 0
c.4x² – 4x + 1 = 0
d.
x² + 2x + 1 = 0
e.3x² – 2x + 3 = 0
f.2x² – 5x + 2 = 0
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quadratic equations
5. For what values of m will the roots of x² – 6x + m = 0 be real?
6. The equation 2x² + kx + 18 = 0 has equal roots.
Find two possible values of k.
Check your answers.
remember!
The discriminant b² – 4ac determines the nature of the roots of ax² + bx + c = 0.
1.If b² – 4ac > 0, there are two real roots.
If b² – 4ac = m², where m
Q, the roots are rational.
Otherwise, they are irrational.
2.If b² – 4ac = 0, there is one rational root.
3.If b² – 4ac < 0, there are no real roots.
3C
For which equations from the list 1 – 7 could you find:
a. two distinct solutions?
b. two equal solutions?
c. no real solutions?
1. x² + 5x + 3 = 0
2.4x² + 9x – 4 = 0
3. x² – 3x – 2 = 0
4.7x² – 3x = 1
5. x² – 6x + 5 = 0
6.4x² + 12x + 9 = 0
7.3y² = y – 1
8. Which equations have real solutions?
9. Which equation has no real solution?
10.Which equations have rational solutions?
11.Which equations have irrational solutions?
Check your answers.
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quadratic equations
discriminants and quadratic graphs
You will now learn how the graphs of quadratic functions illustrate the solutions to quadratic
equations. This work combines the following topics:
• the solutions to quadratic equations (use of quadratic equation)
• the nature of the roots (the discriminant)
• the graphs of quadratic functions (graphing).
discovery exercise
1. Answer the following questions, for the function f(x)=x² – 4x + 3
=
(x – 1)(x – 3)
a. For the general quadratic equation ax² + bx + c = 0 the discriminant is given by b² – 4ac.
Find the discriminant of the equation f(x) = 0.
b. Solve the quadratic equation f(x) = 0.
c. Give the point(s), if any, where the graph of f intersects the x-axis.
d. Sketch the graph of the function f.
Check your answers.
2. Answer the following questions, for the function g(x)=x² – 4x + 4
=
(x – 2)²
a. Find the discriminant of g(x) = 0.
b.Solve g(x) = 0.
c. Give the point(s), if any, where the graph of g intersects the x-axis.
d. Sketch the graph of g.
3. Answer the following questions, for the function h(x) = x² – 4x + 5
a. Find the discriminant of h(x) = 0.
b.Solve h(x) = 0.
c. Give the point(s), if any, where the graph of h intersects the x-axis.
d. Sketch the graph of h.
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quadratic equations
4. Copy and complete the table from your answers to questions 1, 2 and 3.
Function
Intersection
point(s) on
x-axis
Sign of
b² – 4ac
(+, 0, -)
f(x)
(1, 0), (3, 0)
+
Solutions
of y = 0
x
Number of
real roots
of y = 0
2
{1, 3}
g (x)
h (x)
Look for patterns in the table above.
• Where on the graph would you find the solution to y = 0?
• What is the connection between the number of real roots and the number of times
the parabola cuts the x-axis?
• What is the connection between the discriminant and the number of times the parabola
cuts the x-axis?
Copy and complete the following statements to summarise your findings.
a. The roots of the equation ax² + bx + c = 0 are the
of the points where
the graph of the function y = ax² + bx + c intersects the.
b. i. If the graph of y = ax² + bx + c crosses the x-axis twice, then ax² + bx + c = 0
has
real roots and b² – 4ac0.
ii. If the graph of y = ax² + bx + c touches the x-axis, then ax² + bx + c = 0
has
real roots and b² – 4ac0.
iii.If the graph of y = ax² + bx + c does not intersect the x-axis, then ax² + bx + c = 0
has
real roots and b² – 4ac0.
Check your answers.
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quadratic equations
remember!
The points where the graph of y = ax² + bx + c crosses the x-axis are the
points where y = 0; that is, where ax² + bx + c = 0. So the x-values of these points
are the roots of the equation ax² + bx + c = 0.
If the graph crosses the x-axis
twice, then ax² + bx + c = 0
has two real unequal roots; that is,
the x-values of the two points where
the graph crosses the x-axis.
This occurs when b² – 4ac > 0.
If the graph touches the x-axis,
then ax² + bx + c = 0 has two
equal real roots; that is, the x-value
of the point where the graph
touches the x-axis. (This is
sometimes called a repeated
root.) This occurs when
b² – 4ac = 0.
If the graph does not cross the
x-axis, then ax² + bx + c = 0 has
no real roots. This occurs when
b² – 4ac < 0.
y
0
x
0
x
y
y
0
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quadratic equations
3D
1. This graph shows a quadratic function f.
f (x)
a. Use the graph to solve f(x) = 0.
b. Is the discriminant of f(x) = 0 positive,
negative or zero?
-1
0
3
x
-3
2. The graph shows a quadratic function g.
a. What is the solution set of g(x) = 0?
b. Describe the roots of g(x) = 0.
g(x)
0
x
c. What is the sign of the
discriminant of g(x) = 0?
multichoice
Write down the letter that corresponds to the answer you think is correct.
3. Which best describes the nature of the roots of 3x² – 4x + 2 = 0?
a. The roots are not real.
b. There are two distinct real roots.
c. There is only one real root.
d. Both roots are irrational.
e. There are two distinct rational roots.
Question 4 refers to the following graphs.
3
3
-4
0
f (x)
f (x)
f (x)
f (x)
4
x
0
x
-4
0 x
-3
3
0
4
x
I. II. III.IV.
4. The equation f (x) = 0 has two real roots for which functions?
a.
I onlyb.
IV onlyc.
II and IIId.
I and IV
Check your answers.
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4
problems involving quadratic equations
learning outcomes
Form, use and solve linear and quadratic equations.
Determine the nature of the roots of a quadratic equation.
learning intentions
In this lesson you will learn to:
•• form and solve quadratic equations using the quadratic formula
•• determine the nature of the roots of a quadratic equation and apply the discriminant
to find solutions.
introduction
Some algebraic problems need to be solved using quadratic equations. It is important that you
check your answers with the conditions stated in the problem, as quadratic equations
always have two solutions and one of these may be nonsensical. A negative length for
example, or a negative age or a fraction of a person, does not make much sense.
Example 1
Twice the square of a certain whole number is 91 more than 19 times that certain whole number.
Find the number.
Let k be that number.
2k² = 19k Twice the
square of
the number.
19 times
the original
number.
+ 91
91 more
than.
2k² – 19k – 91 = 0
(2k + 7)(k – 13) = 0
Either 2k + 7 = 0 or k – 13 = 0
−
7
k = or k = 13
2
−
Since the original number was a whole number, k = 7 is not a solution to the problem.
2
CheckIf k = 132k² = 2 × 13²
=338
19k + 91 = 91 × 13 + 91
=338
The number required is 13.
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problems involving linear equations
Example 2
A rectangular lawn measures 14 metres by 11 metres.
A path of uniform width surrounds it.
If the area of the path is 186 m2, find the width of the path.
A diagram shows the problem more clearly.
14
11
x
x
Let the width of the path be x metres.
The area of the path is the difference between the areas of the outer and inner rectangles.
Area of outer rectangle = (11 + 2x)(14 + 2x)
Area of inner rectangle = 11 × 14
(2x + 11)(2x + 14) – 11 × 14 = 186
4x² + 50x + 154 – 154 = 186
4x² + 50x – 186 = 0
2x² + 25x – 93 = 0
(2x + 31)(x – 3) = 0
−
31
Either
x = or x = 3
2
Make it simpler by
dividing through
by a common factor.
Since x = –15.5 is not a realistic answer, the required answer is 3.
Check Area of outer rectangle = 17 × 20
=340
Area of inner rectangle = 154
Area of path = 340 – 154
=186
The width of the path is three metres.
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problems involving linear equations
4A
For some of these problems, you may need to use the quadratic formula.
x =
b±
−
b² − 4ac
2a
1. The temperature T, in degrees centigrade, of the element of a toaster is given by
T = 18 + 70x – 4x² (0 < x < 8) where x is the time in seconds since the toaster is turned on.
How long does it take for the temperature of the element to first reach 300°C?
2. The vertical height h, in metres, of a ball thrown upwards by a boy is given by
h = 2 + 20t – 5t² where t is the time in seconds since the ball left the boy’s hand.
At what times is the ball 18 metres above the ground?
Explain why there are two different times.
3. A path of uniform width surrounds a rectangular lawn that measures 25 metres by 16 metres.
If the area of the path is 180 m2, find its width.
4. A man was 26 years old when his son was born, and the product of their ages now is 192.
How old is the son now?
5. Kim is four years older than Kit. The sum of the squares of their ages is 346. Find Kit’s age.
6. The lengths of three sides of a triangle in centimetres are x + 7, x – 2 and 3x, as shown
in this diagram.
a. Use the Theorem of Pythagoras to
find an equation linking
the lengths of the sides.
b. Solve the equation for x.
c. Which value for x is the correct solution?
d. What are the lengths of each side of the triangle?
x+7
x−2
3x
Check your answers.
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problems involving simultaneous equations
learning outcome
Solve simultaneous linear and non-linear equations.
learning intentions
In this lesson you will learn to:
•• form, use and solve linear and quadratic equations
•• solve simultaneous equations.
introduction
The problems you will encounter in this lesson are solved using simultaneous equations.
You will need to find two linear equations to describe each situation, as there are two unknowns
in each situation. This means you will have to introduce two variables and state what each
represents. Then follow the basic problem-solving techniques that you’ve used in the
last two lessons.
Example 1
The sum of two numbers is 23 and their difference is 27.
What are the two numbers?
1. Read the problem carefully.
Think: ‘sumʼ means adding and ‛differenceʼ means subtracting.
2. Consider drawing a diagram.
There’s no point this time.
3. Choose letters for both unknown numbers and state what they represent.
Let the numbers be a and b.
4. Form two equations from the information in the question.
a + b = 23 The sum is 23
a – b = 27 The difference is 27
5. Solve the pair of equations simultaneously.
a + b
=
23(A)
a – b=
27(B)
a + a + b – b = 23 + 27 Adding (A) and (B)
2a = 50
a =25
25 + b=
23Substitute into (B).
–
b=2
6. Check your answer with the conditions stated in the problem.
25 + –2 = 23
25 – –2 = 27
7. Write a sentence answering the question.
The numbers are 25 and –2.
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problems involving simultaneous equations
In example 1, you did not have to distinguish between a and b because you were finding two
numbers. In example 2, it is important to state clearly what each variable represents.
Example 2
Concert tickets cost $110 for the main
floor and $165 for the balcony.
2 000 tickets were sold. The takings were
$260 700. How many balcony tickets
were sold and how many main-floor
tickets were sold?
iStock
You may like to try solving this problem
yourself using the guidelines shown in
example 1. If not, work through the
following questions, which will help
you learn the problem-solving process.
1. What are you asked to find?
2. Choose a letter to represent the number of balcony seats sold and write
a sentence stating this. Repeat for the number of main-floor tickets.
3. Write a mathematical phrase for the takings from balcony tickets.
Write a mathematical phrase for the takings from main-floor tickets.
Write an equation that connects the statements for the takings of the balcony tickets
and the main-floor tickets with the total takings.
4. What information in the question have you not yet used?
5. Write an equation connecting the number of balcony tickets and main-floor tickets sold
with the total number of tickets sold.
6. Solve simultaneously the two equations you have found.
7. Write a sentence answering the original question.
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problems involving simultaneous equations
Answer to example 2
1. The number of balcony tickets and the number of main-floor tickets sold.
2.Let x represent the number of balcony tickets sold.
Let y represent the number of main-floor tickets sold.
3.Balcony ticket takings: 165x
Main-floor ticket takings: 110y
165x + 110y = 260 700
4. 2 000 tickets were sold.
5. x + y = 2 000
6.
165x + 110y = 260 700 (A)
x + y
=
2 000(B)
From (B) y = 2 000 – x
Substitute 2 000 – x for y in (A)
165x + 110(2 000 – x) = 260 700
165x + 220 000 – 110x = 260 700
55x + 220 000 = 260 700
x =740
Put 740 for x in (B)
740 + y
y
Check
x + y
1260 + 740
165x + 110y
165 × 740 + 110 × 1 260
7. Answer:
= 2 000
= 1 260
= 2 000
= 2 000
(Number of tickets)
= 260 700
= 260 700
(Takings)
740 balcony tickets and 1 260 main-floor tickets were sold.
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problems involving simultaneous equations
Use a pair of simultaneous equations to solve each of the following problems.
Remember! The variable(s) you
define must correctly express
what you’re trying to find.
1. Find two numbers such that three times the first added to twice the second makes 30,
and twice the first added to the second makes 17.
2. I’m thinking of two numbers. They differ by seven, and if I add twice the greater number to five times the smaller number I get 42.
3. Half the sum of two numbers is –1, while half their difference is 3 .
2
2
Find the numbers.
4. A man was 32 years old when his son was born and the sum of their ages is now 56.
How old are they both now?
5. Thirty-seven teenagers are going on a picnic. If there are nine more boys than girls,
how many boys and how many girls are there?
6. Kelly weighs 3 kg less than Terri, and together their weights are still 5 kg less than Joe’s.
If Joe weighs 90 kg, what does Kelly weigh and what does Terri weigh?
iStock
7. Ian paid $7.15 for five nectarines and three peaches.
Hemi paid $8.60 for four nectarines and six peaches.
What is the average price of a nectarine and a peach?
iStock
5A
Check your answers.
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problems involving simultaneous equations
simultaneous equations: one linear, one non-linear
The pairs of simultaneous equations you have solved so far have contained two linear
equations, and you could use either the elimination or substitution methods to solve them
algebraically.
For solving pairs of simultaneous equations where only one equation is linear, you must use
substitution. You substitute an expression for x or y obtained from the linear equation into the
non-linear equation.
The non-linear equation can take a variety of forms; for example, y = x² + 5x + 6 and
xy = 12, x² + y² = 4, but the equation you obtain after substitution will be a quadratic equation.
You can solve this by either factorising, or by using the quadratic formula.
Example 3
Solve simultaneously y=x²(A)
y=x + 2 (B)
Start by substituting x + 2 for y in equation (A).
x + 2 = x²
Look closely at this equation. It contains an x² term, so it is a quadratic equation.
The first step in solving quadratic equations is to move all terms on to the same side.
0=
x² – x – 2
x² – x – 2 = 0
Now factorise the left-hand side and find values for x.
(x – 2)(x + 1) = 0
Either x – 2 = 0 or x + 1 = 0
x = 2
x = –1
Continue by substituting each x value into equation (B) to find the y values.
x = 2 gives y = 2 + 2 and x = –1 gives y = –1 + 2
= 4= 1
Check your solutions in equation (A):
4 = 2² so (2, 4) is a solution
1 = (–1)² so (–1, 1) is a solution.
The solutions are: (x, y)
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problems involving simultaneous equations
Example 4
Solve these simultaneous equations.
x² + y² =
13(A)
x – y + 1=
0(B)
x=y – 1(C)Make x the subject of the formula in (B).
(y – 1)² + y²
=
13Substitute for x in (A).
2y² – 2y – 12 = 0 Simplify.
y² – y – 6
=
0Divide by common factor 2.
(y – 3)(y + 2)
=
0Factorise.
Either y − 3 = 0
or y + 2 = 0
Solve.
y = 3
If y = 3, then
x=
3 – 1Substitute into (C) to find x.
y = 2
−
=
2
If y = –2, then
x=–2 – 1
–
=
3
Check in (A): 2² + 3² = 4 + 9
Check both answers using other equation.
=
13
(–3)² + (–2)² = 9 + 4
=
13
Solution is:
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{(2, 3), (–3, –2)} Write out solution.
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problems involving simultaneous equations
5B
Solve the following pairs of simultaneous equations algebraically.
1.
y=x² + 3x – 2
y=x – 3
x² + y²= 10
2.
y =3x + 10
3.
2x – y + 1 = 0
y=x² + x – 5
4. x² + y² =25
x + y =7
5.
xy =15
4x – y =7
Check your answers.
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problems involving simultaneous equations
problems involving simultaneous equations:
one linear, one non-linear
To solve these problems, you will use the same basic techniques that you used earlier.
As there are two unknowns in each problem, you will need to define and use two variables.
Example 5
The difference of two positive numbers is three and the sum of their squares is 65.
What are the numbers?
1. Read the problem carefully.
Think: ‛differenceʼ means subtract and ‛sumʼ means add.
2. Consider drawing a diagram.
There’s no point this time.
3. Choose letters for both unknown numbers and state what they represent.
Let the numbers be a and b.
4. Form two equations from the information in the question.
a – b = 3The difference is 3
a² + b² = 65The sum of the squares is 65
5. Solve the equations simultaneously.
a² + b² = 65 (A)
a – b=3(B)
From (B):
a
In (A): (b + 3)² + b²
b² + 6b + 9 + b²
2b² + 6b – 56
b² + 3b – 28
(b + 7)(b – 4)
Either b+7
b =
=
=
=
=
=
=
=
b+3
65
65
0
0
0
0 or b – 4 = 0
–
7 b = 4
If b = –7, then a = –7 + 3
=−4
If b = 4, then
a = 4+3
= 7
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problems involving simultaneous equations
6. Check your solutions with the conditions stated in the problem.
The question says the numbers are positive, so the negative answers must be eliminated.
7 – 4 = 3 7² + 4² = 49 + 16
=
65 
7. Write a sentence answering the question.
The two positive numbers are 7 and 4.
Example 6
A rectangular paddock has a perimeter of 60 metres and an area of 216 m2.
Find the paddock’s dimensions.
1. Read the problem carefully.
Think how you find the area and perimeter of rectangles.
2. Consider drawing a diagram.
y
x
3. Choose letters for both unknowns and state what they represent.
Let the length of the paddock be x metres and let the width be y metres.
4. Form two equations from the information in the question
2(x + y) =
60The perimeter is 60 m.
xy = 216The area is 216 m2.
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problems involving simultaneous equations
5. Solve the equations simultaneously.
2(x + y)
=
60(A)
xy=
216(B)
From (A):
x + y =30
y = 30 – x
In (B):
x(30 – x) =216
30x – x² =216
x² – 30x + 216 = 0
(x – 12)(x – 18) = 0
Either x – 12 = 0 or x – 18 = 0
x = 12
x = 18
If x = 12, y = 30 – 12
= 18
If x = 18, y = 30 – 18
= 12
6. Check the solutions with the conditions stated in the problem.
Area = 18 × 12
= 216
12
Perimeter =
2(18 + 12)
=
2(30)
=
60
18
7. Write a sentence answering the question.
The dimensions of the paddock are 18 metres by 12 metres.
Don't forget to
include the units.
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problems involving simultaneous equations
5C
Use a pair of simultaneous equations to solve each of the following problems.
1. The sum of two numbers is 25 and their product is 144.
What are the numbers?
2. The difference of two positive numbers is five and the sum of their squares is 97. Find the numbers.
3. A rectangle has a perimeter of 76 centimetres and an area of 325 cm².
Find its dimensions.
4. The hypotenuse of a right-angled triangle is 51 centimetres long and the perimeter
of the triangle is 120 centimetres. Find the lengths of the two shorter sides.
Check your answers.
points of intersection of lines with curves
You can find or illustrate the solutions to simultaneous equations like those you solved in the
last lesson with graphs. The linear equations will produce a straight line; the non-linear
equations a curve.
The coordinates of the point (or points) of intersection of the line and the curve are the solutions
to the simultaneous equations being solved.
5D
The solutions to two pairs of simultaneous equations are illustrated below.
1. Use the graphs to write down:
i. the pair of simultaneous equations involved.
ii. the solution of the pair of equations.
a.b.
y
y
5
3
3
y=3
x² + y² = 25
y = (x + 3)(1 − x)
0
3
−
2
−
1
−
0
1
x
4
3
−
5
x
y = −2x + 5
2. Verify your answers to question 1 (ii) by solving the equations algebraically.
Check your answers.
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6
review activity
learning outcomes
Manipulate algebraic expressions.
Determine the nature of the roots of a quadratic equation.
Form, use and solve linear and quadratic equations.
Solve simultaneous linear and non-linear equations.
learning intentions
In this lesson you will learn to:
•• review work on linear and quadratic equations
•• review work on simultaneous equations
•• answer questions for your teacher to assess.
1. Solve these pairs of simultaneous equations:
a.x – 2y =
7b.
2x + 5y = 34
y = 2x + 223x + 4y = 30
2. Express each of the following using algebraic symbols.
a. Two more than five times a certain number.
b. Six less than twice a certain number.
c. The mean of two numbers.
d. The number of minutes in a certain number of hours.
3. Find two consecutive even integers such that twice the smaller exceeds the larger by 18.
4. A Parent Teacher Association knows it sold a
total of 300 tickets to a picnic and that the proceeds
from the sale of the tickets was $1 080. Unfortunately,
no record was kept of the separate number of adults’
and children's tickets sold. This information is now
needed because a special treat is planned for the children.
If the adults’ tickets cost $5 each and children’s tickets
cost $3 each, solve the PTA’s problem.
5. The hypotenuse of a right-angled triangle is 34 cm long.
Find the lengths of the other two sides if one side is 15 cm longer than the other.
iStock
6A
6. Solve the following pair of simultaneous equations algebraically.
x² + y² = 4
y = x – 2
7. Find the coordinates of the points of intersection of y = –5x + 1 and xy = –6.
8. Show that the line x + 3 = 0 is a tangent to the curve x² + y² = 9.
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review activity
9. The graph shows that the line y = 12 – x cuts the parabola y = x².
a. What pair of simultaneous equations
could be solved using this information?
b. Write down the solution to your equations in (a).
y
16
15
14
13
c. What quadratic equation could be solved using
this information?
12
11
10 y = 12 − x
9
y = x²
8
7
6
5
3
2
1
-
4
-
2
1
2
3
x
10. A piece of wire 56 centimetres long is bent to form a right-angled triangle with a
hypotenuse 25 centimetres long. Find the lengths of the other two sides.
Check your answers.
When you check your answers to this review activity, you may discover that, although your
answers, were correct, you used a different method. Many problems can be solved in more than
one way. We’ve shown only one, and it’s not necessarily the best or easiest method.
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review activity
This must be your own work.
You must sign the cover sheet to confirm this.
This is an open-book assessment task, so you may refer back to any of your mathematical
resources.
You may find this formula useful: x=
–b ± b2 – 4ac
2ac
Show full working steps throughout.
1. The diagram shows a photograph
(shaded) mounted on cardboard.
The cardboard measures 100 mm
by 150 mm, and the border around
the photograph is a uniform x mm wide.
Which is the expression (in mm2)
for the area of the photograph?
a.
b.
c.
d.
e.
150 mm
100 mm
6B
x
x
150 × 100 – 150x – 100x
100 × 150 – 4x2
100 × 150 – 4x
(100 + 2x)(150 + 2x)
(100 – 2x)(150 – 2x)
2. Solve the simultaneous equations:
3x + 2y = 11
2x – 3y = 16
3. Find algebraically, the point(s) of intersection of the line
y = x + 6 and the curve y = x2 + 3x – 2.
4. When a pistol which fires flares is fired vertically on the moon, the height h of the flare
above the surface is given by h = 2 + 60 t – 0.85 t2 metres where t is the time in seconds
after firing the flare.
At what time does the flare reach a height of 50 metres as it rises above the surface
of the moon?
5. Find the value of the discriminant and state the number of real root(s) of the quadratic equation x2 + 4x – 1 = 0.
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review activity
6. For each of the following,define suitable variable(s) and form appropriate equation(s).
Do not solve the equations.
a. I think of a number and decrease it by seven. The result is the same as dividing the
original number by eight.
b. In a netball game Julie scored six more goals than Dianne.
Together they scored 26 goals.
c. I’m thinking of two numbers. They differ by 19. Twice the greater number added to
the smaller number is 59.
7. The length of a rectangular swimming pool is twice its width. The pool is surrounded by a concrete path one metre wide. If the area of the path is 46 square metres,
find the dimensions of the swimming pool.
Show clearly the strategies you use to solve this problem.
8. An advertisement offers spring bulbs for sale. For $20, I could buy either 30 tulip bulbs
and 20 daffodil bulbs, or 20 tulip bulbs and 40 daffodil bulbs.
Model this situation and use your model to find the price of one tulip bulb and the price
of one daffodil bulb.
9. An airline collects a total of $11 858 for tickets from the 80 passengers on an airflight.
Each child passenger pays $55 for their flight ticket. There are 20 children on the flight. Each adult passenger either buys a standard ticket for $198 or a thrifty ticket for $132
for the flight.
How many adult passengers have a thrifty ticket?
10.A rectangular pig pen has perimeter 39.8 metres and area 54.7 square metres.
Find its dimensions.
Your teacher will assess this work.
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answer guide
7
1. equation-solving techniques
1A
1.2(x – 5) = 5x + 2
2x – 10 = 5x + 2
2x – 5x = 2 + 10
–3x =12
x=–4
x
x
3. =
–6
8
2
x
x
= 8 × 2 – 8 × 6
8
x =4x – 48
8×
x – 4x=–48
2. 3(3x + 2) – 2(x – 3) = 5
9x + 6 – 2x + 6 = 5
7x + 12 = 5
7x=–7
x=–1
4.x² – 5x + 6 = 0
(x – 2)(x – 3) = 0
Either x – 2 = 0 or x – 3 = 0
x = 2 or x= 3
x {2, 3}
–3x=–48
x =16
5.
x² =3x
x² – 3x =0
x(x – 3) = 0
Eitherx = 0 or x – 3 = 0
x = 3
x {0, 3}
7.x + 12 = x²
x² – x – 12 = 0
(x – 4)(x + 3) = 0
Either x – 4 = 0 or x + 3 = 0
x=
4 x = –3
x {–3, 4}
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6. 3x² =7x + 6
3x² – 7x – 6 = 0
(3x + 2)(x – 3) = 0
Either 3x + 2 = 0 or x – 3 = 0
2
x=− x = 3
3
2
x {– 3 , 3}
© te ah o o te k u ra p ou n a mu
answer guide
1B
1. y = 3x – 3
y = –2x + 7
2. y = 2x + 6 (A)
2y + x =
7(B)
(A)
(B)
Comparing with y = mx + c,
in (A), m = 3, c = –3
in (B), m = –2, c = 7.
y
y = 3x − 3
Comparing (A) with y = mx + c,
we have m = 2, c = 6.
For (B), we complete a tableof ordered pairs:
x
1
3
5
y
3
2
1
6
y
8
4
6
2
-
2
0
-
2
4
x
x
y = 2x + 7
-
4
-
0
2
-
1C
2y + x = 7
2
2
–
y = 2x + 6
Solution is (2, 3).
1.4x – 3y = 7 (A)
2x + 3y=–1 (B)
6x =6
x =1
In (A), 4 – 3y =7
y=–1
Solution is (1, –1).
2
4
6
2
Solution is (–1, 4).
2.
a.7x – 2y =3 (A)
3x + 8y=–43(B)
28x – 8y = 12 (A) × 4
31x=–31
x=–1
–7 – 2y =3
y=–5
Solution is (–1, –5).
b. 2x + y = 10 (A)
3x + 5y = 29 (B)
–10x – 5y=–50 (A) × –5
–
–7x=
21
x =3
6 + y =10
y =4
Solution is (3, 4).
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answer guide
2.
c.2x – 3y =1
3x – 4y =7
6x – 9y = 3
–6x + 8y=–14
–y = –11
y = 11
In (A):2x − 33 = 1
x = 17
Solution is (17, 11).
1D
(A)
(B)
(A) × 3
(B) × –2
d.3x + 2y=
13(A)
7x + 3y=
27(B)
9x + 6y = 39 (A) × 3
–14x – 6y=–54 (B) × –2
–5x = –15
x =3
In (A): 9 + 2y =13
y =2
Solution is (3, 2).
1.
a.y =4x + 1
y=x + 4
x + 4 = 4x + 1
x – 4x = 1 – 4
–3x=–3
x =1
y = 1 + 4
y =5
Solution is (1, 5).
b.y =2x – 3
y =4x + 5
4x + 5 = 2x – 3
4x – 2x=–3 – 5
2x=–8
x=–4
y = 4 × –4 + 5
y=–16 + 5
y=–11
Solution is (–4, –11).
2
a.2x + 3y =1
y =2x – 5
2x + 3(2x – 5) = 1
2x + 6x – 15 = 1
8x =16
x =2
y = 2 × 2 – 5
y=–1
Solution is (2, –1).
b.2x – 3y =11
y =3x – 6
2x – 3(3x – 6) = 11
2x – 9x + 18 = 11
–
–7x=
7
x =1
y = 3 × 1 – 6
y = –3
Solution is (1, –3).
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answer guide
c.3x + 4y =31
2x=y + 6
2x – 6 = y
3x + 4(2x – 6) = 31
3x + 8x – 24 = 31
11x =55
x =5
2 × 5=
y+6
10=
y+6
y =4
Solution is (5, 4).
d.6x – 3y=–8
x + 2y =7
x=–2y + 7
6(–2y + 7) – 3y=–8
–
12y + 42 – 3y=–8
–
15y + 42 = –8
–15y=–50
50
y =
15
1
y = 3 3
1
x + 2 × 3 3 = 7
x + 6 23 = 7
1
x= 3
1
1
Solution is ( 3 , 3 3 ).
2. translating words into mathematics
2A
1
–
1.a.2xb.
x
2
x
c.d. x + 3
4
e.
x – 5
2B
f. 5 – x
g.2x + 3h. 2(x + 3)
i. x−5
2
j.
x
+3
7
2.a.x + 7b.
7b
c. 6 – md.
2g
e.2x + 2
1
f. a or
a
2
2
g.2h + 3h. x – 4
i.
1
y
or yj.
2p – 5
3
3
1.a.z – 2 = 4
2.a.3x + 7 = 42
b.
q + 6 = 10
b.(x – 5)² = 36
c.
m + 4 = 9
c.
x + 6 = 3x
d.7n = 21 or n × 7 = 21
d.
x + 1x = 4.25
e. 17 – x = 4
e.2x – x² = –32
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answer guide
3. a. i. Let the numbers be x and y.
ii.
x + y = 42
x – y = 3
b. i. Let the numbers be m and n.
ii.
m + n = 27
mn = 50
c. i. Let the numbers be u and v.
ii.u + v = 4
1 1
+
= –0.3
u v
2C
1.a.2mb.
m–5
d. m – 7
c.2m + 3
2. a. i. Let the remaining distance be x km.
ii.
15 + x = 34 or 34 – x = 15
b. i. Let the bottom mark be m.
ii.
4 m = 48
c. i. Let the number of children be c.
ii.
3c + 3 = 24
d. i. Let the postal code be x.
ii.
x = 5 × 500 + 5
e. i. Let the width of the rectangle
be x metres.
f. i. Let the number of sheep Charlie sold be x.
ii.
George sells x + 50 sheep
x + x + 50 = 160
x
18
ii.
x + 18 + x + 18 = 56
or 2(x + 18) = 56
or 2x + 36 = 56
2D
1. Let the width of the rectangle be x metres.
x
2x
x + 2x + x + 2x= 96
6x= 96
x= 16
56
2. Let the first of the numbers be m.
m + (m + 1) + (m + 2) + (m + 3) + (m + 4) = 725
5m + 10 = 725
5m = 715
m= 143
The first number is 143.
Check 143 + 144 + 145 + 146 + 147 = 725
The lawn’s width is 16 m.
Check 16 + 32 + 16 + 32 = 96
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answer guide
3. Let the price of the cake be k cents.
Price of pie is 2k
Price of cola is k – 40
(k – 40) + k + 2k =880
4k – 40 = 880
4k =920
k =230
Check Cost of pie, cake and cola
= 2.30 + 4.60 + 1.90
=
8.80
The price of the cake was $2.30.
4. a. Let the speed for the first two hours be x km/h
b. In the first two hours he travels 2x km
c. Speed for final three hours is (x – 15) km/h
d. In these three hours he travels 3(x – 15) km
e.2x + 3(x – 15) = 480
f.
2x + 3x – 45 = 480
5x =525
x =105
g. The motorist travelled at 105 km/h for the
first 2 hours.
h.
Check Travels 210 km in first 2 hours.
Travels for 3 hours at 90 km/h;
that is 270 km, so, total distance
is 480 km.
5. a. Let the present age of the younger
6. Let the size of the smaller angle be mº.
tree be x years.
The other angle is (2m – 3)°
i. So present age of older tree is
7x years.
m°
(2m − 3)°
ii. In five yearsʼ time the age of
the younger tree is x + 5 years.
iii.In five yearsʼ time, the age of the m + 2m – 3 = 180
older tree is 7x + 5 years.
3m= 183
b.7x + 5 = 4(x + 5)
m= 61
c.7x + 5 = 4x + 20
The smaller angle is 61°.
7x – 4x = 20 – 5
Check The larger angle = 2 × 61 – 3
3x = 15
=
122 – 3
x = 5
=
119°
The younger tree is five years old now.
119° + 61° = 180°
Check In 5 yearsʼ time, the younger
tree will be 10 years old.
The older tree is now 35 years
old, so in five years it will
be 40 years old. This is four
times the age of the younger tree.
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answer guide
7. Let the number of years JJ has belonged to the club be x.
JJ has paid $336 subscription for x – 3 years.
Total = entrance fee + sub for first 3 years + sub for
spentremaining years
2 099=
35 + 720 + 336(x – 3)
2 099 = 755 + 336x – 1 008
–
2 099=
253 + 336x
2 352=
336x
7 = x
JJ has belonged to the club for 10 years.
CheckEntrance fee 35
Sub for 3 years at $240 per year
720
Sub for 7 years at $192 per year 1 344
Total$2 099
3. quadratic equations
3A
1. x² + 5x + 3 = 0
1x² + 5x + 3 = 0
a = 1, b = 5, c = 3
x =
x =
x =
x =
x =
x =
58
b±
can be written as
b² − 4ac
2a
−
5 ± 5² − 4 × 1 × 3
2×1
−
5 ± 25 − 12
2
−
5±
13
−
2
5 + 13
2
−
0.70
–
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or
x =
5−
13
−
2.4x² + 9x – 4 = 0
a = 4, b = 9, c = –4
−
9 ± 9² − 4 × 4 × −4
x= Using the formula.
2×4
x =
x =
9±
81 + 64
8
9±
145
−
−
8
9 + 145
8
−
x =
x = 0.38
or x =
9−
145
−
8
or x = –2.63 (2 d.p.)
2
or x = –4.30 (2 d.p.)
© te ah o o te k u ra p ou n a mu
answer guide
3. x² – 3x – 2 = 0
a = 1, b = –3, c = –2
b±
4.7s² – 3s = 1
7s² – 3s – 1 = 0
a = 7, b = –3, c = –1
b² − 4ac
2a
−
x =
x =
x =
x =
x =
x = 3.56
( 3) ± (−3)² − 4 × 1 × −2
2×1
3±
3±
x =
x =
x =
x =
s =
s =
s =
s = 0.65
17
2
3 + 17
3 − 17
or x =
2
2
x =
( 6) ±
6 ± 36 − 20
2
6 ± 16
2
But
3±
37
14
3 + 37
14
3−
or s =
s =
37
14
0.22 (2 d.p.)
–
6.4x² = –12x – 9
4x² + 12x + 9 = 0
a = 4, b = 12, c = 9
or
x =
6−4
2
10
2
x =
2
2
12 ± (12)² − 4 × 4 × 9
2×4
−
x =
x =
x =
x =
x = –1
16 is exactly 4
6±4
2
Either
6+4
x = 2
9 + 28
14
0.56 (2 d.p.)
(−6)² − 4 × 1 × 5
2×1
− −
3±
–
5. x² – 6x + 5 = 0
a = 1, b = –6, c = 5
s =
9+8
2
−(−3) ± (−3)² − 4 × 7 × −1
2×7
− −
Rearranging
x =
x = 5
Note: The solutions for this quadratic equation
could have been found by factorising
x² − 6x + 5 = (x − 5)(x − 1)
12 ± 144 − 144
−
8
12 ± 0
8
−
12
8
−
1
2
x = 1
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answer guide
3B
2.Since b² – 4ac = 0, there is one repeated rational root.
1. Since the discriminant is less than 0,
the equation has no real roots.
3. The discriminant is greater than 0 and not the square of a rational number, so there are two irrational roots.
4. a.
x² + 3x – 10 = 0
1x² + 3x + –10 = 0
a = 1, b = 3, c = –10
b² – 4ac = 3² – 4 × 1 × –10
= 9 + 40
= 49
b² – 4ac > 0 and b² – 4ac = 7² (7
Q).
There are two real roots, both of
which are rational.
b. x² + 3x – 11 = 0
1x² + 3x + –11 = 0
a = 1, b = 3, c = –11
b² – 4ac = 3² – 4 × 1 × –11
= 9 + 44
= 53
b² – 4ac > 0 and b² – 4ac is not the square
of a rational number.
There are two real roots, both of which
are irrational.
c. 4x² – 4x + 1 = 0
a = 4, b = –4, c = 1
b² – 4ac = (–4)² – 4 × 4 × 1
=
16 – 16
=
0
Since b² – 4ac = 0, there is one
rational root.
d.x² + 2x + 1 = 0
a = 1, b = 2, c = 1
b² – 4ac = 2² – 4 × 1 × 1
=
0
Since b² – 4ac = 0, there is one
rational root.
e.3x² – 2x + 3 = 0
a = 3, b = –2, c = 3
b² – 4ac = (–2)² – 4 × 3 × 3
=
4 – 36
–
=
32
Since b² – 4ac < 0, there are
no real roots.
f.
2x² – 5x + 2 = 0
a = 2, b = –5, c = 2
b² – 4ac = (–5)² – 4 × 2 × 2
= 9
b² – 4ac > 0 and b² – 4ac = 3² (3
Q).
There are two real roots, both of
which are rational.
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© te ah o o te k u ra p ou n a mu
answer guide
3C
5. The roots will be real if b² – 4ac ≥ 0
x² – 6x + m = 0
a = 1, b = –6, c = m
(–6)² – 4 × 1 × m ≥ 0
36 – 4 m ≥ 0
36 ≥ 4 m
9 ≥ m
The roots are real if m ≤ 9.
6.
2x² + kx + 18 = 0
a = 2, b = k, c = 18
The roots are equal if b² – 4ac = 0
k² – 4 × 2 × 18 = 0
k² = 144
k = ±12
The roots are equal if k = ± 12.
1. x2 + 5x + 3 = 0
a = 1, b = 5, c = 3
b2 – 4ac = 52 – 4 × 1 × 3
= 13
b2 – 4ac > 0 → two distinct solutions
2. 4x2 + 9x – 4 = 0
a = 4, b = 9, c = -4
b2 – 4ac = 92 – 4 × 4 × -4
= 145
b2 – 4ac > 0 → two distinct solutions
3. x2 – 3x – 2 = 0
a = 1, b = -3, c = -2
b2 – 4ac = (-3)2 – 4 × 1 × -2
= 17
b2 – 4ac > 0 → two distinct solutions
4. 7x2 – 3x = 1
7x2 – 3x – 1 = 0
a = 7, b = -3, c = -1
b2 – 4ac = (-3)2 – 4 × 7 × -1
= 37
b2 – 4ac > 0 → two distinct solutions
5. x2 – 6x + 5 =0
a = 1, b = - 6, c = 5
b2 – 4ac = (-6)2 – 4 × 1 × 5
= 16
b2 – 4ac > 0 → two distinct solutions
6. 4x2 + 12x + 9 = 0
a = 4, b = 12, c = 9
b2 – 4ac = 122 – 4 × 4 × 9
= 0
b2 – 4ac = 0 → two equal solutions
7. 3y2 = y – 1
3y2 – y + 1 = 0
a = 3, b = -1, c = 1
b2 – 4ac = (-1)2 – 4 × 3 × 1
= -11
b2 – 4ac < 0 → no real solutions
8.
9.
10. Equations 5 and 6 have rational solutions
[b2 – 4ac = 0 or a square number]
Equation 7 has no real solutions
Equations 1 to 6 have real solutions
11. Equations 1 to 4 have irrational solutions
[b2 – 4ac ≠ 0 or a square number]
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answer guide
discovery exercise
1.a.x² – 4x + 3 = 0
a = 1, b = –4, c = 3
b² – 4ac = (–4)² – 4 × 1 × 3
= 16 – 12
=
4
b. f (x) = 0
(x – 1)(x – 3) = 0
x {1, 3}
c. The graph cuts the x-axis at (1, 0)
and (3, 0).
d. f (x) = x² – 4x + 3
f (0) = 3
The graph cuts the y-axis at (0, 3).
The axis of symmetry is x = 2
f (2) = 2² – 4 × 2 + 3
–
= 1
The vertex is (2, –1).
y
2
0
-2
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2
4
6 x
f(x)=(x 1)(x 3)
x=2
© te ah o o te k u ra p ou n a mu
answer guide
b.g(x) = 0
(x – 2)² = 0
x {2}
2.a. x² – 4x + 4 = 0
a = 1, b = –4, c = 4
b² – 4ac = (–4)² – 4 × 1 × 4
= 16 – 16
= 0
c. The graph touches the x-axis at (2, 0).
y
x=2
4
2
0
g(x) = ( x
2
2
2)
4
x
3.a. x² – 4x + 5 = 0
a = 1, b = –4, c = 5
b² – 4ac = (–4)² – 4 × 1 × 5
= 16 – 20
–
= 4
d.g(x) = x² – 4x + 4
g(0) = 4
The graph cuts the y-axis at (0, 4).
The axis of symmetry is x = 2
so the vertex is at (2, 0).
b.
h(x) = 0 has no real solution since the discriminant is negative.
c. The graph does not intersect the x-axis.
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answer guide
d.h(x) = x² – 4x + 5
h(0) = 5
The graph cuts the y-axis at (0, 5).
y
6
4
If h(x) = 5 5 = x² – 4x + 5
0 =
x² – 4x
0 =
x(x – 4)
x {0, 4}
A second point on the graph is (4, 5).
The axis of symmetry is x = 2
h(2) = 2² – 4 × 2 + 5
= 1
The vertex is at (2, 1).
4.
x=2
2
h(x) = x2 – 4x + 5
0
2
4
6 x
Function
Intersection point(s)
on x-axis
Sign of
b² = 4ac
(+, 0, -)
f(x)
(1, 0), (3, 0)
+
x
{1, 3}
2
g(x)
(2, 0)
0
x
{2}
1
h(x)
none
-
x
{ }
0
a. The roots of the equation
ax² + bx + c = 0 are the x-values
of the points where the graph of
the function y = ax² + bx + c
intersects the x-axis.
Solutions
of y = 0
Number of
real roots
of y = 0
b. i. If the graph of y = ax² + bx + c crosses the x-axis twice,
then ax² + bx + c = 0
has two real roots and b² – 4ac > 0.
ii. If the graph of y = ax² + bx + c touches the x-axis,
then ax² + bx + c = 0
has one real root and b² – 4ac = 0.
iii.If the graph of y = ax² + bx + c
does not intersect the x-axis,
then ax² + bx + c = 0 has no
real roots and b² – 4ac < 0.
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answer guide
3D
1. a. The parabola cuts the x-axis at –1 and 3 so
f(x) = 0
x {–1, 3}
b. Since there are two real solutions,
the discriminant is positive.
2. a. The parabola does not cut the x-axis so the
solution set is the empty set { } or Ø.
b.
g(x) has no real roots.
c. The discriminant has a negative sign.
multichoice
3.3x2 – 4x + 2 = 0
a = 3, b = –4, c = 2
b2 – 4ac = (–4)2 – 4 × 3 × 2
–
= 8
Answer: a. (the roots are not real)
4. If the equation has two real roots,
then the parabola will cut the x-axis
at two distinct points.
Answer: c
(II and III)
4. problems involving quadratic equations
4A
1. When the temperature is 300°C:
300 = 18 + 70x – 4x²
so 4x² – 70x + 282 = 0
a = 4, b = –70, c = 282
−
b ± b² − 4ac
x = 2a
x =
( 70) ±
− −
70 ±
The quadratic formula.
(−70)² − 4 × 4 × 282
2×4
388
=
= 11.21 or 6.29 (2 d.p.)
Because 0 < x < 8, the value 11.21
can be eliminated. So the time to first
reach 300°C is 6.29 seconds (2 d.p.).
8
© te ah o o t e k ur a p o un a m u
2. When the ball is 18 m above the ground
18= 2 + 20t – 5t²
0=5t² – 20t + 16
a = 5, b = –20, c = 16
− −
( 20) ± (−20)² − 4 × 5 × 16
t= 2×5
Using the quadratic formula.
t =
20 ±
80
10
= 2.89 or 1.11 (2 d.p.)
The ball is 18 m above the ground at
1.11 seconds (2 d.p.) and 2.89 seconds
(2 d.p.). There are two different times when the ball can be 18 metres above
the ground as the ball can be rising
or falling.
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answer guide
3. Let the width of the path be x metres
Area of outer rectangle = (16 + 2x)(25 + 2x)
Area of inner rectangle = 16 × 25
(2x + 16)(2x + 25) – 16 × 25
4x² + 50x + 32x + 400 – 400
4x² + 82x – 180
2x² + 41x – 90
=
=
=
=
180
180
0
0
(2x + 45)(x – 2) = 0
Either 2x + 45 = 0
or x – 2 = 0
x = –22.5 or
x = 2
4. Let the son’s present age be n years.
Father’s age is n + 26 years
n(n + 26) = 192
n² + 26n – 192 = 0
(n + 32)(n – 6) = 0
Either n + 32 = 0 or n – 6 = 0
n = –32 or
n = 6
Since n = –32 is not a realistic answer,
the son’s present age is 6 years.
Check If the son is now 6, his father is 32.
The product of 6 and 32 is 192.
x
16
25
x
Since x = –22.5 is not a realistic answer for
a width, the width of the path is 2 metres.
Check Area outer rectangle = 20 × 29
=
580 m²
Area inner rectangle
=
400 m²
Area path
=
580 – 400
=
180 m²
5. Let Kit’s age be k years. Kim’s age is k + 4
k² + (k + 4)² = 346
k² + k² + 8k + 16 = 346
2k² + 8k – 330 = 0
k² + 4k – 165 = 0
(k – 11)(k + 15) = 0
Either k – 11 = 0 or k + 15 = 0
k = 11 or k = –15
Since –15 is not a realistic age,
Kit’s age is 11 years.
Check If Kit is 11, Kim is 15
11² + 15² = 121 + 225
= 346
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© te ah o o te k u ra p ou n a mu
answer guide
6.a.(3x)² + (x – 2)² = (x + 7)²
b. 9x² + x² – 4x + 4 = x² + 14x + 49
9x² – 18x – 45 = 0
a = 9, b = –18, c = –45
x =
b±
−
b² − 4ac
2a
− −
( 18) ± (−18)² − 4 × 9 × −45
=
2×9
18 ±
=
1944
18
x = 3.45 or –1.45 (2 d.p.)
c.If x = –1.45 then 2x and x – 2 have
negative values, which are not realistic
lengths for the sides. When x = 3.45
the sides 2x and x – 2 have positive values.
Hence x = 3.45 is the correct solution.
d. The sides are
3x = 10.35 cm
x – 2 = 1.45 cm x + 7 = 10.45 cm
(2 d.p.)
(2 d.p.)
(2 d.p.)
10.45 cm
1.45 cm
10.35 cm
5. problems involving simultaneous equations
5A
1. Let the numbers be x and y.
3x + 2y = 30 (A)
2x + y = 17 (B)
From (B):
y = 17 – 2x
In (A): 3x + 2(17 – 2x) =30
3x + 34 – 4x =30
–x=–4
x =4
In (B):
2 × 4 + y =17
y =9
(x , y) = (4, 9)
Check In (A): 3 × 4 + 2 × 9 = 30
© te ah o o t e k ur a p o un a m u
2. Let the greater number be x and the
smaller number be y.
x – y = 7 (A)
2x + 5y = 42 (B)
From (A):
x =7+y
In (B): 2(7 + y) + 5y = 42
14 + 2y + 5y = 42
7y = 28
y = 4
In (A): x – 4 = 7
x = 11
(x, y) = (11, 4)
Check In (B): 2 × 11 + 5 × 4 = 42
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answer guide
3. Let the two numbers be x and y.
1
(x + y)= 2
1
(x – y)
2
–
1
(A)
2
3
= 2 (B)
Multiply both equations by 2 to
remove fractions
x + y = –1(C)
x – y
= 3 (B)
Add:
2x = 2
x =1
1 + y = –1
In (C):
y = –2
(x, y) = (1, –2)
1
1
Check In (B): 2 (1 – –2) = 2 × 3
3
=2
4. Let the man’s age now be x years and his son’s age now be y years.
x = y + 32 (A)
x + y = 56 (B)
Substitute y + 32 for x in (B):
y + 32 + y = 56
2y = 24
y = 12
In (A): x = 12 + 32
=44
The man’s present age is 44
and his son’s age is 12.
Check Sum of ages = 44 + 12
= 56
Man’s age when
son born =
44 – 12
=
32
5. Let the number of boys be b and
the number of girls be g.
b + g = 37 (A)
b = g + 9 (B)
Substitute g + 9 for b in (A):
g + 9 + g = 37
2g = 28
g = 14
6. Let Kelly’s weight be k kg
and Terri’s weight be t kg.
k = t – 3 (A)
k + t + 5 = 90 (B)
In (B): b = 14 + 9
b= 23
In (A): k = 44 – 3
= 41
There are 14 girls and 23 boys on the picnic.
Check 14 + 23 = 37 (people on picnic)
23 – 14 = 9 (9 more boys than girls)
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Substitute t – 3 for k in (B)
t – 3 + t + 5 = 90
2t = 88
t = 44
Kelly weighs 41 kg and Terri
weighs 44 kg.
Check Kelly weighs 3 kg less than Terri. Combined weight is
85 kg, which is 5 kg less
than 90 kg.
© te ah o o te k u ra p ou n a mu
answer guide
7. Let the average price of a nectarine be
n cents and the average price of a peach
be p cents.
5n + 3p = 715 (A)
4n + 6p = 860 (B)
–10n – 6p = –1 430 (A) × –2
Add: –6n = –570
n = 95
In (A):
5 × 95 + 3p = 715
3p = 240
p = 80
The average price of a nectarine is 90 cents
and the average price of a peach is 80 cents.
Note: In the last two questions, the
variables must be defined for the weight and price, respectively. ‘Let Kelly be k’
or ‘The nectarine is n’ are not sufficiently well defined.
5B
1.y = x² + 3x – 2 (A)
y = x – 3
(B)
x – 3 = x² + 3x – 2
0 =
x² + 2x + 1
0 =
(x + 1)(x + 1)
x = –1
If x – 1, then y = –1 – 3
–
= 4
Check in (A): (–1)² + 3 × –1 – 2 = 1 – 3 – 2
–
= 4
–
–
Solution is: (x, y)
{( 1, 4)}
© te ah o o t e k ur a p o un a m u
2. x² + y² = 10(A)
y = 3x + 10 (B)
x² + (3x + 10)² = 10
x² + 9x² + 60x + 100 = 10
10x² + 60x + 90 = 0
x² + 6x + 9 = 0
(x + 3)² = 0
x = –3
If x = –3, then y = 3 × –3 + 10
= 1
Check in (A):
Solution:(x, y)
(–3)² + 1² = 10
{(–3, 1)}.
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answer guide
3.y = x² + x – 5 (A)
2x – y + 1 = 0 (B)
From (B): y = 2x + 1
2x + 1 = x² + x – 5
0 = x² – x – 6
0 = (x – 3)(x + 2)
Either x – 3 = 0 or x + 2 = 0
x = 3
x = –2
If x = 3, y = 2 × 3 + 1 = 7
If x = –2, y = 2 × –2 + 1 = –3
Check In (A): (3, 7)
gives 3² + 3 – 5 = 7
In (A): (–2, –3)
gives (–2)2 + –2 – 5 = –3
Solution:(x, y)
{(3, 7), (–2, –3)}.
5.
xy = 15
4x – y = 7
From (B): y = 4x – 7
x(4x – 7) = 15
4x² – 7x – 15 = 0
(4x + 5)(x – 3) = 0
Either 4.x² + y² = 25 (A)
x + y = 7 (B)
From (B):
y = 7–x
x² + (7 – x)² = 25
x² + 49 – 14x + x² = 25
2x² – 14x + 24 = 0
x² – 7x + 12 = 0
(x – 3)(x – 4) = 0
Either x – 3 = 0 or x – 4 = 0
x = 3
x = 4
If x = 3, y = 7 – 3 = 4
If x = 4, y = 7 – 4 = 3
Check in (A) 3² + 4² = 9 + 16
= 25✓
Solution is: (x, y)
{(3, 4), (4, 3)}.
(A)
(B)
4x + 5 = 0 or x – 3 = 0
5
x = – 4
5
x = – 4,
x = 3
5
y = 4 × – 4 – 7 = –12
If If x = 3, y = 4 × 3 – 7 = 5
5
Check in (A) – 4 × –12 = 15
70
3 × 5 = 15 ✓
Solution is: (x, y)
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{(– 5 , –12), (3, 5)}.
4
© te ah o o te k u ra p ou n a mu
answer guide
5C
1. Let the numbers be x and y.
• The sum is 25 x + y = 25 (A)
• The product is 144 xy = 144 (B)
From (A): y = 25 – x
x(25 – x) = 144
25x – x² = 144
0 = x² – 25x + 144
0 = (x – 16)(x – 9)
Either x – 16 = 0 or x − 9 = 0
x = 16 or
x=9
If x = 16, y = 9
If x = 9, y = 16
The numbers are 9 and 16.
Check 16 × 9 = 144
Solution is: (x, y)
{(16, 9), (9, 16)}.
3. Let the dimensions be
x centimetres and y centimetres.
2. Let the numbers be x and y.
• The difference is 5
x – y = 5 (A)
• Sum of squares is 97 x² + y² = 97 (B)
From (A): x = y + 5
(y + 5)² + y² = 97
y² + 10y + 25 + y² = 97
2y² + 10y – 72 = 0
y² + 5y – 36 = 0
(y + 9)(y – 4) = 0
Either y + 9 = 0 or y – 4 = 0
y = –9 or
y=4
–
–
If y = 9, x = 4
If y = 4, x = 9
Since the numbers are positive,
the numbers are 9 and 4.
Check 9 – 4 = 5
9² + 42 = 81 + 16 = 97
Solution is:
(x, y)
{(–4, –9), (9, 4)}.
x
y
• The area is 325 cm²
xy = 325 (A)
• The perimeter is 76 cm2x + 2y = 76 (B)
From (B): 2x = 76 – 2y
x = 38 – y
y(38 – y) = 325
38y – y² = 325
0 =
y² – 38y + 325
0 =
(y – 13)(y – 25)
Either y – 13 = 0 or y – 25 = 0
y = 13
y = 25
If y = 13, x = 25
If y = 25, x = 13
Check Area = 25 × 13 = 325
Perimeter = 2 × 13 + 2 × 25 = 76
The dimensions of the rectangle are 25 centimetres
by 13 centimetres.
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answer guide
4. Let the lengths of the sides be x centimetres
and y centimetres.
51 cm
y
x
• The perimeter is 120 cm x + y + 51
• The hypotenuse is 51x² + y² From (A): x (69 – y)² + y² 4761 – 138y + 2y² 2y² – 138y + 2 160 y² – 69y + 1 080 (y – 24)(y – 45) = 120 (A)
= 51² (B)
= 69 – y
= 512
= 2 601
=0
=0
=0
Either y – 24 = 0 or y – 45 = 0
y = 24
y = 45
If y = 24, x = 45
If y = 45, x = 24
Check In (B):
24² + 45² = 576 + 2 025
= 2 601
= 51²
5D
The lengths of the two other sides are
24 centimetres and 45 centimetres.
1. a. i.
y = (x + 3)(1 – x)
y = 3
ii. (0, 3), (–2, 3)
b. i. x² + y² = 25
y = –2x + 5
ii. (0, 5), (4, –3)
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answer guide
2. a. Here is the algebraic solution
for graph (a).
y = (x + 3)(1 – x)(A)
y =
3(B)
3 =
(x + 3)(1 – x)
– 2
3 =
x – 2x + 3
x² + 2x = 0
x(x + 2) = 0
Either x = 0 or x + 2 = 0
x = –2
Solution is: (x, y)
{(0, 3), (–2, 3)}.
b. Here is the algebraic solution
for graph (b).
x² + y²=
25(A)
y = –2x + 5 (B)
x² + (–2x + 5)² = 25
x² + 4x² – 20x + 25 = 25
5x² – 20x = 0
5x(x – 4) = 0
Either 5x = 0 or x = 0 or
x–4=0
x=4
If x = 0, then y = –2 × 0 + 5 = 5
If x = 4, then y = –2 × 4 + 5 = –3
Solution is: (x, y)
{(0, 5), (4, –3)}.
6. review activity
6A
1.
a. x – 2y y
x – 2(2x + 22) x – 4x – 44 –3x x y y
=7
= 2x + 22
=7
=7
= 51
= –17
= 2 × –17 + 22
= –12
Solution is (–17, –12).
© te ah o o t e k ur a p o un a m u
b.
2x + 5y 3x + 4y –
8x – 20y 15x + 20y 7x x 2 × 2 + 5y 5y y Solution is (2, 6).
= 34
= 30
= –136
= 150
= 14
=2
= 34
= 30
=6
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(A)
(B)
(A) × –4
(B) × 5
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answer guide
2. You must state clearly what each
letter represents.
a.Let x be the number 2 + 5x
b.Let x be the number2x – 6
x+y
c. Let the two numbers be x and y
2
d.Let t be the number of hours 60t
3. Let the smaller integer be x
The next even integer is x + 2
2x – 18 = x + 2
x = 20
The two integers are 20 and 22.
4. Let the number of adults’ tickets sold be a.
Let the number of children’s tickets sold be c.
a + c = 300 (A)
5a + 3c = 1 080 (B)
Add: −3a – 3c = −900
2a= 180
5. Let the length of the shorter side be x cm
In (A):
90 + c = 300
c = 210
x² + x² + 30x + 225 = 1 156
2x² + 30x – 931 = 0
210 children’s and 90 adults’ tickets
were sold.
Check
Number of tickets sold = 90 + 210
= 300
Proceeds from sales = 90 × 5 + 210 × 3
= $1 080
Check 2 × 20 – 18 = 40 – 18
= 22
34 cm
x
x + 15
x² + (x + 15)2 = 34²
Pythagoras
a = 2, b = 30, c = −931
x =
x =
b±
−
b² − 4ac
2a
30 ± 30² − 4 × 2 × −931
2×2
−
= 15.34 or –30.34 (2 d.p.)
Since a negative length is unrealistic, the value of x is 15.34 only.
Hence the lengths of the two sides are:
x = 15.3 cm (1 d.p.)
and x + 15 = 15.34 + 15
= 30.3 cm (1 d.p.)
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answer guide
6.
x² + y² = 4 (A)
y = x – 2 (B)
x² + (x – 2)² = 4
x² + (x² – 4x + 4) = 4
x² + x² – 4x = 0
2x² – 4x = 0
x² – 2x = 0
x(x – 2) = 0
Either x = 0 or x – 2 = 0
x = 2
If x = 0, y = 0 – 2y = –2
If x = 2, y = 2 – 2 y = 0
Check In (A) (0, –2) gives 0² + (–2)² = 4
In (A) (2, 0)gives 2² + 0² = 4
Solution is: (x, y)
{(0, –2), (2, 0)}.
7.y = –5x + 1 (A)
xy = –6 (B)
Substitute –5x + 1 for y in (B):
x(–5x + 1) = –6
–
5x² + x + 6 = 0
5x² – x – 6 = 0
(5x – 6)(x + 1) = 0
Either 5x – 6 = 0 or x + 1 = 0
x = 6
x = –1
5
6
6
–
If x =
, y= 5×
+ 1 = –5
5
5
If x = –1, y = –5 × –1 + 1 = 6
Check In (B): (
6 –
, 5) gives × –5 = –6
5
In (B): (–1, 6) gives –1 × 6 = –6
6
The points of intersection are ( , –5)
5
and (–1, 6)
8. x² + y² = 9
x + 3 = 0
From (B): x = –3
–
Substitute 3 for x in (A):
(–3)² + y² = 9
y² = 0
y = 0
(A)
(B)
Solution is: (−3, 0).
Since the line intersects the circle at
one point only, the line x = –3 is a tangent.
© te ah o o t e k ur a p o un a m u
9.a.y = x²
y = 12 – x
b.(x, y)
{(–4, 16), (3, 9)}
c.x² = 12 – x
x² + x – 12 = 0
The solutions to this equation are x {−4, 3}.
They are the x-coordinates of the points
of intersection.
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75
answer guide
10. Let the lengths of the sides be
x centimetres and y centimetres.
25 cm
y
x
• The wire is 56 cm long
x + y + 25 = 56 (A)
• It forms a right-angled triangle x² + y² = 25² (B)
From (A): x = 31 – y
In (B): (31 – y)² + y² = 25²
961 – 62y + y² + y2 = 625
2y² – 62y + 336 = 0
y² – 31y + 168 = 0
(y – 7)(y – 24) = 0
Either y – 7 = 0 or y – 24 = 0
y = 7
y = 24
If y = 7, x = 31 – 7 x = 24
If y = 24, x = 31 – 24 x = 7
Check In (B): 24² + 7² = 576 + 49
= 625
= 25²
The lengths of the other two sides of the
triangle are 24 centimetres and 7 centimetres.
76
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© te ah o o te k u ra p ou n a mu
appendix
derivation of the quadratic formula
Consider the general equation:
ax² + bx + c =
b
c
Divide through by a to make
x² + x + a
a
the coefficient of x² equal to 1.
Move the constant term to the other side.
=
0
b
x=
a
−
b
b
x+(
)² =
a
2a
−
x² +
0
c
a
Complete the square for
b
1
b
x by adding (
×
)²
a
2
a
b
that is, (
)² to both sides.
2a
x² +
x² +
c
b
+(
)²
a
2a
Write the left-hand side as a perfect square and simplify the square on the right.
(x +
b
)² =
2a
−
Add the fractions on the right-hand side.
(x +
b
)² =
2a
−
(x +
−
4ac
b²
b
)² = 4a² + 4a²
2a
(x +
b² − 4ac
b
)² =
4a²
2a
Take the square root of both sides.
x+
b
=
2a
Simplify the square root using the rule
x+
b
2a
=
x+
b
2a
=
p
=
q
p
q
Move
b
to the right-hand side.
2a
Write the result as a single fraction.
© te ah o o t e k ur a p o un a m u
x =
x =
c
b²
+
a
4a²
c × 4a
b²
+
a × 4a
4a²
b² − 4ac
4a²
±
±
b² − 4ac
4a²
± b² − 4ac
2a
b
±
2a
b² − 4ac
2a
b±
b² − 4ac
2a
−
−
MX2062
77
acknowledgements
Every effort has been made to acknowledge and contact copyright holders. Te Aho o Te Kura Pounamu apologises for any
omissions and welcomes more accurate information.
Graph diagrams created using GeoGebra program from http://www.geogebra.org/cms. Used in any medium for education
and its promotion by permission.
All other illustrations copyrighted to Te Aho o Te Kura Pounamu, Wellington, NZ.
Photos:
istock: Solving problems – 15798783, School picnic – 14466457,
Theatre – 15377330, Peaches – 16017618, Tangerines – 16643212.
78
MX2062
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