Download THE QUADRATIC QUANDARY

THE QUADRATIC QUANDARY
By Jay Snyder
[email protected]
I
What is a Quadratic Equation?
II
What is a Quadratic Used For?
III
Solve the Quadratic by Factoring
IV Quadratic Formula
V
Application
I
What is a Quadratic Equation?
“quad” means square. The exponent is 2 (squared). The equation has degree 2.
Standard Form (generic form for parabola):
ax2 + bx + c = 0
where a, b & c are numbers and aǂ0.
Identify the quadratic :
Linear Equation : 3x  2  17
Graph is a line
Quadratic Equation : 2 x 2  3x - 14  0
Cubic Equation : 4 x3  2 x 2  3x  2  0
Graph is a parabola
Graph is a " squiggle"
Examples of quadratics:
x 2  8 x  12  0
Parabola:
Circle :
Ellipse :
x2  y2  9
x2 y2

1
9
4
Hyperbola :
x2 y2

1
16 9
II
What is a Quadratic Used For?
Quadratics are widely used in science, business, and engineering.
1. The U-shape of a parabola can describe the trajectories of an object when it flies
through the air. From the equation we can find the height of an object after a given
length of time or tell how long it will take for an object to reach a certain height. The
military can predict when or where artillery will hit the earth or target.
2. Quadratics can be used to design parabolic shapes such as satellite dishes and car
headlights. Most of the objects we use every day were created by applying a quadratic
function.
3. The shape of cables in a suspension bridge can be determined with the quadratic.
4. Quadratics are used to help forecast business profit & loss and assist in determining
minimum and maximum values.
5. Quadratic equations are used in situations where two things are multiplied together and
they both depend of the same variable. For example, when working with area, if both
dimensions are written in terms of the same variable, a quadratic equation can be used.
6. The quadratic can be used to explain how planets in our solar system revolve
(elliptically) around the sun.
Gravity
Sales
Area of Frame
Best price
area
height
profit
width
time
price
III
Solve the Quadratic by Factoring
A
Factoring
1 GCF
2 Riddle Game
3 Difference of 2 Squares
4 Factoring Combinations
5 Grouping & Riddle Game “Expanded”
[jmath.net/Algebra 1 course/video #17]
B
Solve by Factoring
1 The Zero Guarantee
2 Solve the Quadratic Equation with 3 Steps
a) Set Equation Equal to Zero
b) Factor
c) Use the Zero Guarantee
Factoring
GCF (Greatest Common Factor)
1) Define factor as a noun: numbers multiplied together to get a given product.
Factors of 12: 1 x 12, 2 x 6, 3 x 4
{1, 2, 3, 4, 6, 12}
Factors of 18: 1 x 18, 2 x 9, 3 x 6
{1, 2, 3, 6, 9, 18}
2) CHANGE THE MINDSET FOR ALGEBRA!
What numbers can you divide by? These are the factors.
What numbers can you divide into 12? 1, 2, 3, 4, 6, 12
NOTE: If you can divide by a number, it must have another factor.
12 is divisible by 6. Therefore, there is another factor (which is 2).
3) GCF of 12 & 18
Common Factors of 12 & 18:
1 2 3 4 6 12
1 2 3 6 9 18
GCF
The GCF of 12 & 18 is 6.
(The largest number that will divide into 12 & 18)
4) Prime Factorization (factor tree)
Prime numbers have exactly 2 factors.
7 is a prime number. Its factors are 1 & 7.
19 is a prime number. Its factors are 1 & 19.
Prime factorization breaks a number down using its factors until all the factors
are prime.
12
12
2 x 6
3 x 4
2 x 3
2 x 2
2 x 2 x 3 = 12 OR 22 x 3
2 x 2 x 3 = 12 OR 22 x 3
5) Define factor as a verb: to express as a product of 2 or more quantities.
We factor to change addition or subtraction to multiplication!
Arithmetic:
10 + 15
The GCF of 10 & 15 is 5.
Use the GCF (5) and factor (pull out the GCF)
and change to multiplication.
5(2 + 3)
The addition (10+15) has been changed to multiplication 5(2+3)!
Algebra:
10x + 15
5(2x + 3)
Algebra:
GCF = 5
Addition changed to multiplication!
9x2 + 12x
To find GCF, find GCF of the numbers (coefficients) first (3)
Then find the GCF of the variables (x). GCF = 3x
3x(3x + 4)
Addition changed to multiplication!
Riddle Game
Example 1:
I’m thinking of 2 numbers whose product is 12 and whose sum is 8.
What are the 2 numbers?
Answer: 2 and 6
Example 2:
I’m thinking of 2 numbers whose product is 24 and whose sum is 11.
What are the 2 numbers?
Answer: 3 and 8
Example 3:
I’m thinking of 2 numbers whose product is -15 and whose sum is 2.
What are the 2 numbers?
Answer: 5 and -3
Example 4:
I’m thinking of 2 numbers whose product is 12 and whose sum is -7.
What are the 2 numbers?
Answer: -3 and -4
Review Distributive Property:
3(x + 4) = 3x + 12
x(x +2) = x2 + 2x
(x + 3)(x + 2) = x2 + 2x + 3x + 6 = x2 + 5x + 6
To factor a quadratic trinomial, play the riddle game.
Example 1:
+
x
x2 + 5x + 6
Riddle Game: Multiply to get 6 and add to get 5.
Answer: 2 and 3
( x + 2)(x + 3)
Answer to riddle
NOTE: the addition has been changed to multiplication.
+
x
2
Example 2: Factor x + 9x + 14
Riddle Game: Multiply to get 14 and add to get 9.
Answer to riddle: 2 and 7
Answer: (x + 2)(x + 7)
***The riddle game will not work unless the x2 coefficient is 1.
3x2 + 4x – 6 will not factor with the riddle game because of the coefficient 3.
Difference of 2 Squares
Difference refers to subtraction.
A square number is the product of a number times itself.
Examples of square numbers:
1, 4, 9, 16, 25, 36, 49, 64, x2, 9x2, …
Look at this multiplication:
(x - 4)(x + 4) = x2 + 4x – 4x – 16 = x2 - 16
Notice 4x – 4x cancels out the x term.
The final answer is a difference of 2 squares (the x2 and 42).
Therefore, factoring the difference of 2 squares is one of the easiest expressions to
factor (as long as you recognize it! It needs to JUMP OFF THE PAGE AT YOU!!!) It’s
easier than finding the GCF or using the riddle game.
Examples:
x 2  9  ( x  3)( x  3)
x 2  4  ( x  2)( x  2)
x 2  49  ( x  7)( x  7)
x 2  64  ( x  8)( x  8)
9 x 2  4 y 2  (3x  2 y )(3x  2 y )
4 x 6  25  (2 x 3  5)( 2 x  5)
Factoring Combinations
Always look for the GCF before other factoring tools.
Example 1:
Example 2:
Example 3:
Factor 8x2 – 32
8(x2 – 4)
(GCF = 8)
8(x + 4)(x – 4)
(Difference of 2 Squares)
Factor 3x2 + 3x – 18
3(x2 + x – 6)
(GCF = 3)
3(x + 3)(x – 2)
(Riddle Game)
Factor 2x4 – 2
2(x4 – 1)
(GCF = 2)
2(x2 + 1)(x2 - 1)
(Difference of 2 Squares)
2(x2 + 1)(x + 1)(x – 1) (Difference of 2 Squares)
Solve by Factoring
The Zero Guarantee (Zero Product Property)
If a•b = 0, what conclusion can you make about a or b?
a or b must equal 0.
THIS SIMPLE CONCEPT IS USED TO SOLVE QUADRATICS!
To use this concept, we must have a multiplication problem. That’s why we factor.
It changes addition or subtraction to multiplication.
If 8x = 0, then x must equal 0.
If (x – 2)(x – 7) = 0, then x = 2 OR x = 7
If (x + 3)(x – 12) = 0, then x = -3 OR x = 12
If x(x + 4)(x + 5) = 0, then x = 0, x = -4 OR x = -5
Solve the Quadratic Equation with 3 Steps
Example 1: Solve the Quadratic
x2 + 7x + 10 = 0
Step 1: Make sure the equation is set equal to zero. The Zero Guarantee
will not work if the equation does not equal zero.
This equation is already set equal to zero.
Step 2: Factor the equation.
(x + 2)(x + 5) = 0
(Riddle Game)
Step 3: Use the Zero Guarantee to solve.
Solution to equation: {-2,-5}
Example 2: Solve the Quadratic
x2 - 4x - 12 = 0
Step 1: The equation is already set zero.
Step 2: Factor the equation.
(x – 6)(x + 2) = 0
(Riddle Game)
Step 3: Use the Zero Guarantee to solve.
Solution to equation: {-2,6}
Example 3: Solve the Quadratic
5x2 – 45 = 0
Step 1: The equation is already set equal to zero.
Step 2: Factor the equation.
5(x2 – 9) = 0
(GCF)
5(x + 3)(x – 3) = 0
(Difference of 2 Squares)
Step 3: Use the Zero Guarantee to solve.
Solution to equation: {-3,3}
Example 4: Solve for x.
2x2 + 18x = -16
Step 1: Set the equation equal to zero. Add 16 to both sides of equation.
2x2 + 18x + 16 = 0
Step 2: Factor the equation.
2(x2 + 9x + 8) = 0
(GCF)
2(x + 1)(x + 8) = 0
(Riddle Game)
Step 3: Use the Zero Guarantee to solve.
Solution to equation: {-1,-8}
IV
Quadratic Formula
If you can’t or don’t want to factor a quadratic equation, you can solve it with the
Quadratic Formula.
The Quadratic Formula looks a little intimidating. But with a closer look, it’s not!
The Quadratic Formula:
x
 b  b 2  4ac
2a
NOTE: There are 2 a’s, 2 b’s and 1 c. (Highlighting is helpful!)
All we need to do is substitute the numbers for a, b & c
and then “crunch” the numbers!
Identify a, b & c
Standard form of the quadratic (parabola) is: ax2 + bx + c = 0
where a, b and c are the numbers we need for the Quadratic Formula.
Identify a, b and c in the following examples.
Example 1:
Example 2:
Example 3:
Example 4:
Example 5:
4x2 +7x + 3 = 0
x2 + 2x - 5 = 0
6x2 - x + 8 = 0
3x2 + 6x = 0
5x2 - 6 = 0
a=4
a=1
a=6
a=3
a=5
b=7
b=2
b = -1
b=6
b=0
c=3
c = -5
c=8
c=0
c = -6
Evaluate the Formula
1) Explain the  symbol. It is a shortcut for writing two problems at one time.
8  2 means 8 + 2 and 8 – 2 OR 10 and 6.
2) Solve the same equation by factoring and then by using the Quadratic Formula.
Solve for x by factoring
Solve for x with the Quadratic Formula
x2  5x  6  0
x2  5x  6  0
( x  2)( x  3)  0
a  1, b  5, c  6
Solution : {2,3}}
x
 b  b 2  4ac
2a
Plug in a, b and c :
 5  52  4(1)(6)
x
2(1)
x
 5  25  24
2
x
5 1
2
x
 5 1  5 1
,
2
2
x
4 6
,
2 2
x  2,3
Solution : {2,3}
As you can see, factoring to solve is much less work than using the Quadratic Formula.
Always try to factor first!
3) Solve using the Quadratic Formula.
(this equation does not factor)
3x 2  7 x  3  0
a  3, b  7, c  3
x
 b  b 2  4ac
2a
Plug in a, b and c :
 7  7 2  4(3)(3)
x
2(3)
x
 7  49  36
6
x
 7  13
6
  7  13  7  13 
Solution : 
,

6
6


Use of the Discriminant
The discriminant is the expression under the radical sign in the Quadratic Formula:
b2 – 4ac
Evaluating the discriminant will tell you how many solutions the equation has.
It does not tell you what the solutions are!
A discriminant greater than zero means there will be 2 solutions to the equation.
A discriminant that equals zero means there will be 1 solution to the equation.
A discriminant less than zero means there are no solutions to the equation.
Look at these 3 possible solutions:
The discriminant is 5.
There are 2 solutions.
The discriminant is 0.
There is 1 solution.
15  5
7
3 0
7
The discriminant is -7.
There are no real solutions.
2 7
9
V
Application
The quadratic equation ho = -16t2 + vt + hi is used with gravity problems. With this
equation we can find the time an object will reach a given height or find the height of an
object after a given amount of time.
ho = the height of the object at a given time (t)
t = the time in seconds that the object travels
v = the initial velocity of the object in feet per second
hi = the initial height of the object in feet
Example 1:
Carol dove into a swimming pool from a 15 foot high diving board. Her initial upward
velocity was 8 feet per second. Find the time (t) in seconds it took Carol to enter the
water.
Substitute the values from the problem into the equation ho = -16t2 + vt + hi
ho = 0
The height of Carol when she enters the water
v=8
The initial velocity 8 feet per second
hi = 15
Carol’s initial height from a 15 foot diving board
0 = -16t2 + 8t + 15
a = -16
Solve for t using the Quadratic Formula.
b=8
t
 8  82  4(16)(15)
2(16)
t
 8  1024  8  32

 32
 32
t
24  40
,
 32  32
c = 15
t  .75, 1.25
The time (t) that it took Carol to enter the water cannot be negative.
Therefore, it took her 1.25 seconds to enter the water.
Example 2:
Using the previous problem, what was Carol’s height after 1 second?
Substitute 1 for t and solve for ho.
ho = -16(1)2 + 8(1) + 15
ho = 7
After 1 second Carol was 7 feet above the water.
Example 3:
Tom threw a baseball in the air from a height of 6 feet with an initial upward velocity of
14 feet per second. James caught the ball on its way down at a point 4 feet above the
ground. How long was the ball in the air before James caught it?
Substitute the values from the problem into the equation ho = -16t2 + vt + hi
ho = 4
The height of the ball when it was caught
v = 14
The initial velocity 14 feet per second
hi = 6
Height of ball when it left Tom’s hand
4 = -16t2 + 14t + 6
Solve for t using the Quadratic Formula
Set the equation equal to zero
(subtract 4 from both sides of equation)
0 = -16t2 + 14t + 2
Identify a, b and c
a = -16
b = 14
c=2
t
 14  142  4(16)( 2)
2(16)
t
 14  324  14  18

 32
 32
t
4  32
,
 32  32
t  .125, 1
The time (t) the ball was in the air cannot be negative.
Therefore, the ball took 1 second to go from Tom to Jerry.