Solving Quadratic Functions

Lesson 5-3

Objective• Today, you will . . .

• solve quadratic functions by using a variety of methods.

TEKS:b2A,d1A,d3A,d3C,d3D

Some Notes on Quadratic Functions1. The graphs of quadratic functions are

parabolas.

2. The solution(s) that you will be looking for are the x-intercepts of the parabola.

3. The x-intercepts are also called the “roots,” “solutions,” or the “zeros”

4. Quadratic functions can have one, two, or no real solutions.

5. Quadratic functions that have no real solutions have complex (imaginary) solutions.

Different Graphs of Quadratics

x-intercepts, roots, zeros

Two real solutions

X=-6

X=1

Different Graphs of Quadratics

x-intercept, root, zero

One real solutions

X=3

Different Graphs of Quadratics

No x-intercepts (roots or zeros)

No real solutions

No Solution

Today you’ll find…

• The solutions to quadratic equations by factoring

• For example:

GIVEN: y = Ax2 + Bx + C

FIND: The solutions, roots, zeros, or x-intercepts

Solve by factoring: x2 + 9x + 20 = 0

x2 + 9x + 20 = 0

(x + 5)(x + 4) = 0

x + 5 = 0 x + 4 = 0 x = - 5 x = - 4

So, its two roots, solutions, zeros

are -5 & -4

1 x 20

2 x 20

4 x 5

Solve by factoring: x2 = -7x + 18

x2 + 7x - 18 = 0

(x - 2)(x + 9) = 0x - 2 = 0 x + 9 = 0

x = 2 x = - 9 So, its two roots, solutions, zeros

are 2 & -9

1 x 18

2 x 9

3 x 6Hint: The sign of “B” goes with the largest factor!

Solve by factoring: x2 + 100 = 29x

x2 - 29x + 100 = 0

(x - 4)(x - 25) = 0

x - 4 = 0 x - 25 = 0

x = 4 x = 25 So, its two roots, solutions, zeros

are 4 & 25

1 x 100

2 x 50

4 x 25

5 x 20

10 x 10

Solve by factoring: x2 - 9 = 0

x2 - 9 = 0

(x + 3)(x - 3) = 0

x + 3 = 0 x - 3 = 0 x = - 3 x = 3

So, its two roots, solutions, zeros are

-3 & 3

1 x 9

3 x 3

Solve by factoring: x2 + x = 6

x2 + x - 6 = 0

(x - 2)(x + 3) = 0x - 2 = 0 x + 3 = 0 x = 2 x = - 3

So, its two roots, solutions, zeros

are 2 & -3

1 x 6

2 x 3Hint: The sign of “B” goes with the largest factor!

Solve by factoring: x2 - 6x = - 8

x = 2

x = 4

1 x 8

2 x 4

Solve by factoring: x2 - 8 = - 7x

1 x 8

2 x 4

x = 1

x = - 8

Solve by factoring: 2x2 + 13x + 15 = 0

2x2 + 13x + 15 = 0 Multiply AxC = 30

Determine the factors of 30 that give you 13

1 x 30

2 x 15

3 x 10

(x + 3) (x + 10) Write the factors

(x + 3) (x + 10 ) Divide the #’s by A 2 2

(2x + 3) (x + 5) If not divisible, send it in front of “x”,

if divisible then simplify.

2x + 3 = 0 x + 5 = 0 Now solve both factors!

2x = -3 x = -5

x = -3/2

Your solutions are

x = -3/2 and x = -5

Solve by factoring: 3x2 + 16x + 21 = 0

3x2 + 16x + 21 = 0 Multiply AxC = 63

Determine the factors of 63 that give you 16

1 x 63

3 x 21

7 x 9

(x + 7) (x + 9) Write the factors

(x + 7) (x + 9 ) Divide the #’s by A 3 3

(3x + 7) (x + 3) If not divisible, send it in front of “x”,

if divisible then simplify.

3x + 7 = 0 x + 3 = 0 Now solve both factors!

3x = -7 x = -3

x = -7/3

Your solutions are

x = -7/3 and x = -3

Solve by factoring: 2x2 – 5x = 7

2x2 – 5x – 7 = 0 Multiply AxC = -14

Determine the factors of -14 that give you -5

1 x 14

2 x 7

(x + 2) (x – 7) Write the factors

Remember, sign of “B” goes to the largest factor, in this case, the negative goes to the 7.

(x + 2) (x – 7) Divide the #’s by A 2 2

(x + 1) (2x – 7) If not divisible, send it in front of “x”,

if divisible then simplify.

x + 1 = 0 2x – 7 = 0 Now solve both factors!

x = -1 2x = 7

x = 7/2

Your solutions are

x = -1 and x = 7/2

Your turn!Solve by factoring: 5x2 + 4x = 12

5x2 + 4x – 12 = 0 A x C = -60 1 x 60 4 x 15

2 x 30 5 x 12

3 x 20 6 x 10

(x – 6) (x + 10) Write the factors (watch your signs!)

(x – 6) (x + 10) Divide the #’s by A 5 5

(5x – 6)(x + 2) If not divisible, send it in front of “x”,

if divisible then simplify.

5x – 6 = 0 x + 2 = 0 Now solve both factors!

5x = 6 x = -2

x = 6/5Your solutions are

x = 6/5 and x = -2

You try these by factoring . . .

1. 2x2 – 3x = 0

2. 4x2 + 5 = -9x

3. 6x2 + 55x = -9

x = 0 and 3/2

x = -1 and -5/4

x = -9 and -1/6