Download CLIL Module: QUADRATIC EQUATIONS Class: 3AL SY: 2014/2015

CLIL Module: QUADRATIC EQUATIONS
Class: 3AL
SY: 2014/2015
Duration: 10 hours
Teaching/Learning Methodologies:
 learning by doing,
 video-lectures (Khan Academy)
 work in pairs/groups
Written
Said
Scientific Notation
Order of
magnitude
1,256,721 m
Conversion
Gm
31,256,721 ms
s
721,011 L
cl
23,028,000,341
mm
0.34 Gs
hm
ds
21.072 cm3
km3
425.27 dm2
Mm2
0.005 K
GRUPPO N° _________
________________________ ____________________________ __________________________
_______________________________ ____________________________
mK
SCAFFOLDING LANGUAGE: Solving Equation
Vocabulary
…is grater then…
…is greater then or equal to…
…is less then or equal to…
…is less then…
algebraic fraction
canonical/standard/normal form
coefficient
constant term
determinate equation
domain
equation
existence condition
field
identity
impossible equation
indeterminate equation
intermediate coefficient
leading coefficient
left hand side
like terms
linear equation
polynomial of degree one/two
quadratic binomial
quadratic equation
right hand side
root
solution
solution set
solving process
to adjust the numerators
to cancel/to eliminate
to collect like terms
to factorise
to find the solution
to rationalize the denominator
to solve
to work out the value of x
two minuses make a plus
unknown
value
variable
zero product property
to remove/get rid of/ multiply out/ expand the brackets
to isolate x by putting/grouping all the xs on one side of the equation
to rearrange the equation in the standard form
to find the l.c.m. (lowest or least common multiple)
to find the l.c.d. (lowest or least common denominator)
to turn the second fraction upside down/ to take the reciprocal
to simplify by multiplying out the numerator
Example 1
3( 4x+ 6) = -9 - ( 9x- 6)
What do we do first?
We remove/get rid of/multiply out/ expand the brackets.
This gives us: 12x+18 = -9 - 9x+ 6
Why do we put +? What is the rule? Two minuses make a plus.
What’s the next step? We isolate x by putting/grouping all the xs on one side of the equation.
OR: We make x the subject of the equation. OR: We collect like terms. 12x+ 9x = -9 + 6 -18
Now we simplify and we get:
21x = -21.
Now we can work out the value of x. OR: We can arrive at the answer.
Divide both sides by 21.
So… OR: Therefore it follows that… x = -1
Finally, you can check the answer/solution/root is correct by substituting the value of x into the equation.
Task 1
Work with your partner. Practise using the scaffolding language to solve these equations:
1)
5( x+1) - 2 = -1- 3( x- 4)
2)
2x - 3 6x - 2 1+ x
=
10
15
5
3)
Task 2
How would you go about solving these problems?
1) The sum of squares of two consecutive odd natural numbers is 74. Find the numbers.
x +1
x+2
-2 =
x- 2
1- x
Let the first number be x. The second odd number is two more then x and can be written as _________ .
Since, sum of their squares is 74, we have __________________________________________________ .
Remove the brackets ____________________________________________ (*) .
Rearrange to equal zero ___________________________________________________ .
This gives us a _________________________ equation or a polynomial of degree _______ , set equal to zero.
The factor ________ is called the leading coefficient, ________ may be called the intermediate coefficient and – 35
is the ________________ .
Factorise the (*) ________________________________________________________ .
Now solve the equation, putting each bracket equal to zero (= __________________________ property):
Therefore the two roots are __________ and ____________ .
Now, x can not be negative as it is a ______________ number. Hence x = _______ .
So, the numbers are _______ and ___________ .
2) The sum of the squares of two consecutive even numbers is 164. Find the numbers.
3) The length of a rectangular garden is
and breadth of the garden.
7 mmore than its breadth.
If area of the garden is
4) The sum of two natural numbers is 12. If sum of their reciprocals is 9, find the numbers.
Task 3
Solve the following quadratic equations by factorisation:
1)
( 2x+ 3) ( x+ 2) = 0
25x2 -10x+1= 0
2
3) 5x - x = 0
2
4) x - 5x- 6 = 0
2
5) 5x -1= 0
2
6) 3+ x = 0
2
7) 3x - 4x- 7 = 0
2
8) 3+ 4x -8x = 0
2
9) x + x+1 = 0
2)
144 m2 , find the length
KEY WORDS FOR ALGEBRA
Match the words below with the definitions. The first one has been done for you.
expression (solving process) substitute (first side) coefficient constant real
variable
identity eliminate terms (left side)
impossible
expand
simplify
(right side)
(add or subtract)
solve transpose
(multiply or divide)
(second side)
(solution set)
equation
1.
An EXPRESSION is a collection of quantities, made up of constants and variables, connected by signs for
operations and usually without an equals sign.
2.
The ________________ ,in a simple algebraic expression, are the quantities that are linked to each other by +
or – signs.
3.
A _______________ is a value that does not change whenever it is used for a defined purpose. It is usually a
number, but it can also be a letter that represents a certain value.
4.
A __________________ is a symbol that can take any value from a given range of values.
5.
An ________________ is an equality between two algebraic expressions which is true for every numeric
value given to their variables.
6.
__________________ is to rearrange an equation in order to produce an equivalent version of the equation
that is easier to work with.
7.
__________________ is to collect together all the like terms in an algebraic expression so they make a single
term.
8.
The ________________________ of an equation consists of all the values that satisfy the equation.
9.
________________________ an equation means to find its solution set.
10. In the _____________________ of an equation you can ________________ the same quantity to both
sides and you can ___________________ both sides by the same non-zero quantity without altering the
solution set of the equation.
11. _____________________ is to remove the brackets and do as much multiplication as possible in order to
make the equation a collection of terms connected only by + and – signs.
12. An _________________ is an equality between two algebraic expressions which is true for some numeric
values given to their variables.
13. The algebraic expression after the equals sign is called the ___________ or ____________ ; the one before
the equals sign is called the _______________ or _________________ .
14. A ____________________ is the constant in front of a variable or group of variables.
15. __________________ is to remove a variable or a number from an expression.
16. If the solution set of the equation is the empty set then the equation is called ___________________
equation.
17. ___________________ is to replace one expression or part of an expression with something of equivalent
value.
18. An equation is called indeterminate equation if its solution set is the set of the ____________ numbers.
LET US SUM UP
True or False?
SENTENCES
T
F
WHY?
1 5 is not a quadratic equation, but it can be reduced to a quadratic
=
x 2
equation.
x+
x2 + x +1= 0 is not a quadratic equation.
3x2 - 2 + 3x = 0 is a quadratic equation in standard form.
Fill the gaps:
1. An equation of the form
form.
ax2 + bx+ c = 0, a ¹ 0, b, c Î R is called a ______________ equation in ________
2. The value(s) of the variable which satisfy a quadratic equation are called its _________ or ____________.
3. If you can _______________ ax2 + bx+ c = 0, a ¹ 0, b, c Î R, into product of linear factors, then the
roots of the quadratic equation can be obtained by equating each factor to _____________ .
4. Roots of the quadratic equation in standard form are given by the formula ___________________________.
5.
b2 - 4ac is called ________________ of the quadratic equation and it is usually denoted by D .
6. If
D > 0,
then the quadratic equation has ___________________ roots.
7. If _________, then the quadratic equation has two equal (coincident) roots.
8. If
D < 0 , then the quadratic equation has ____________ roots.
SOLVING QUADRATIC EQUATIONS
HOW TO SOLVE QUADRATIC EQUATIONS:
Step 1:
Write equation in Standard Form.
Step 2:
Factor the quadratic equation
Step 3:
Once ( ) are separated, set each ( ) = to 0 and solve for the variable.
Step 4:
Check each of the roots in the ORIGINAL quadratic equation
EXAMPLES: Solve each equation. Check the roots.
1.
4.
r 2  12r  35  0
y 2  3y  28
2.
5.
y 2  11y  24  0
x 2  x  30
3.
x 2  5x  6  0
6.
3x 2  3x  7x  3
SOLVING QUADRATIC EQUATIONS
RECALL SOLVING STEPS:
Step 1:
Step 2:
Step 3:
Step 4:
INCOMPLETE (SPECIAL) QUADRATICS:
ax 2  bx  0 and
ax 2  c  0
Solve the following equations. Check all the roots:
1.
z2  4  0
2.
a2  36  0
3.
x 2  4x
4.
x 2  8x  0
5.
3x 2  12  0
6.
5y 2  45
CHALLENGE PROBLEMS: Solve using the “formula” and check the roots
7.
x(x  3)  40
8.
x
3

5 x2
9.
3x x 2

4
8
QUADRATIC WORD PROBLEMS
CONSECUTIVE INTEGERS
Review of Consecutive Integers “Let Statements”
Integers
Consecutive Integers
Consecutive Even Integers
Consecutive Odd Integers
EXAMPLES:
1.
When the square of a certain number is diminished by 9 times the number, the result is 36. Find the number.
2.
A certain number added to its square is 30. Find the number.
3.
The square of a number exceeds the number by 72. Find the number.
4.
Find two positive numbers whose ratio is 2:3 and whose product is 600.
5.
The product of two consecutive odd integers is 99. Find the integers.
6.
Find two consecutive positive integers such that the square of the first is decreased by 17 equals 4 times the
second.
7.
The ages of three family children can be expressed as consecutive integers. The square of the age of the
youngest child is 4 more than eight times the age of the oldest child. Find the ages of the three children.
8.
Find three consecutive odd integers such that the square of the first increased the product of the other two is
224.
GEOMETRIC WORD PROBLEMS
Remember:
 Draw a picture and Write a “Let” statement
 Write an equation
 Solve the equation ( REMEMBER: YOU CAN’T HAVE A NEGATIVE LENGTH)
 Check to see if your solution makes sense
 Re-Read the problem to make sure you answered the question
EXAMPLES:
1.
The ratio of the measures of the base and the altitude of a parallelogram is 3:4. The area of the
parallelogram is 1,200 square centimeters. Find the measure of the base and altitude of the parallelogram.
2.
The altitude of a triangle is 5 less than its base. The area of the triangle is 42 square inches. Find its base and
altitude.
3.
The length of a rectangle exceeds its width by 4 inches. Find the dimensions of the rectangle it its area is 96
square inches.
4.
If the measure of one side of a square is increased by 2 centimeters and the measure of the adjacent side is
decreased by 2 centimeters, the area of the resulting rectangle is 32 square centimeters. Find the measure of
one side of the square.
5.
Joe’s rectangular garden is 6 meters long and 4 meters wide. He wishes to double the area of his garden by
increasing its length and width by the same amount. Find the number of meters by which each dimension
must be increased.
MORE GEOMETRIC PROBLEMS
1.
In a trapezoid, the smaller base is 3 more than the height, the larger base is 5 less than 3 times the height,
and the area of the trapezoid is 45 square centimeters. Find, in centimeters, the height of the trapezoid.
2.
If the length of one side of a square is tripled and the length of an adjacent side is increased by 10, the
resulting rectangle has an area that is 6 times the area of the original square. Find the length of a side of the
original square.
3.
The length of a rectangle is 7 units more than its width. If the width is doubled and the length is increased by
2, the area is increased by 42 square units. Find the dimensions of the original rectangle.
4.
The side of one square is 2 centimeters longer than the side of the second square. If the sum of their areas is
2
100cm , find the length of the side of each square.
Name: ___________________
11/02/2015
3 AL
TEST
Quadratic Equations
Score: _____ / 100
MARK: ________
Pass Mark: 60/100
æx ö 4 2
ç -1÷ = x , fill in the table below the ‘solving steps’ and the ‘sentences’,
è2 ø 9
2
1) (14 points) Given the equation
putting them in the correct order:
Solving steps:
7x2 + 36x- 36 = 0
x2
4
- x +1 = x2
4
9
9x2 - 36x + 36 16x2
=
36
36
9x2 - 36x+ 36 =16x2
Sentences:
cancel out the common factors
multiply out the brackets
rearrange the equation in the standard form
find the l.c.d. and adjust the numerators
Step
Sentence
1)
2)
3)
4)
Then solve the equation in the standard form following the steps:
1) work out the discriminant
2) find the roots
2) (30 points) Solve each equation by factoring: (In case that CANNOT BE solved by factoring, GIVE REASONS)
a)
x+ x2 = 0
d)
9y2 -12y+ 4 = 0
b)
9y2 -1 = 0
e)
x2 +16 = 0
c)
x2 + x- 30 = 0
f)
y2 - 3y+ 5 = 0
3) Solve the equations:
a) (10 points)
x
1
1
- 2
= 2
x -1 3x + 3x 2x - 2x
2
æ 3ö
æ3 ö æ3 ö
2
b) (10 points) ç x + ÷ - x = ç - x÷ × ç - x÷
è 2ø
è2 ø è2 ø
2
c) (10 points)
1
2
x +1 ( x+1) ( x -1)
=
( x -1) +
5
2
15
d) (5 points)
x2 - 2x- 4 = 0
4) (5 points) Find
k if x2 + 5x+ k -1= 0
has real roots:
5) (8+8=16 points) WORD PROBLEMS. Solve with quadratic equations:
a)
Find two consecutive natural numbers such that the sum of their square is 85.
b) A fenced-in yard, with the dimensions 6 m by 8 m, is expanded by the same distance in each dimension.
2
The resulting region is now 143 m .
How much distance was added to each dimension?
(fenced-in yard: enclosed garden)