quadratic functions and equations - ddtwo

Quadratic Functions and Equations

What is Quadratic Function? Equation?

How to Solve Quadratic Function/ Equation?

Friday, January 31, 2020

Quadratic Function and Equation

A quadratic function is in the standard form y = ax2 + bx + c or f(x) = ax2 + bx + c

A quadratic equation is a quadratic function equated to zero

The standard quadratic equation form is ax2 + bx + c = 0 where a, b, and c are numbers with a 0.

Solving Quadratic Equation

Quadratic Equation can be solve by

Taking the square root

Factoring

Completing the square

Quadratic formula

Graphing calculator

Friday, January 31, 2020

Solving Quadratic Equation

Only Quadratic Equation of the form

ax2 + c = 0

Example:

Which of the following quadratic equation can be solved by taking the square root?

can be solved by taking the square root.

B.) 3x2 – 9x = 0A.) 2x2 + 8 = 0

C.) x2 + 4x – 5 = 0 D.) none of these

Solving Quadratic Equation

Example: Solve by taking the square root:

1.) 2x2 – 8 = 0 2.) (x – 3)2 + 8 = 44

2x2 = 8

2 2

4

x2 = 4

x2 =

x = 2

x = 2 Or x = -2

(x – 3)2 = 36

(x – 3)2 = 36

x – 3 = 6

x = 6 + 3 Or x = -6 + 3

x = 9 Or x = -3

Solving Quadratic Equation

Example: Solve by taking the square root:

3.) 2x2 + 2 = 0 4.) (x – 3)2 + 8 = -28

2x2 = -2

2 2

-1

x2 = -1

x2 =

x = i

x = i Or x = -i

(x – 3)2 = -36

(x – 3)2 = -36

x – 3 = 6i

x = 3 + 6i Or x = 3 – 6i

Numbers with i is called an imaginary number

Real and Imaginary together are called Complex number

Solving Quadratic Equation

Try this! : Solve by taking the square root:

5.) x2 – 8 = 0 6.) (x + 1)2 + 8 = 58

x2 = 8

8x2 =

x = 2

x = 2

Or x = -2

(x + 1)2 = 50

(x + 1)2 = 50

x + 1 = 5

x = -1 + 5

Or x = -1 – 5

2

2

2

2

2

2

Solving Q. E. by Factoring

Example: Factor and use the zero property to solve the Q.E.

1. x² + 3x − 10 = 0

= 0( x ) ( x ) 2 and -5 -3

-2 and 5 3

1 and -10 -9

-1 and 10 9

- 2 + 5

Factors of -10 Sum of Factors

Using zero property

x – 2 = 0 or x + 5 = 0

x = 2 or x = -5

Friday, January 31, 2020

Solving Q. E. by Factoring

Example: Factor and use the zero property to solve the Q.E.

2. x² − 5x + 6 = 0

= 0( x ) ( x ) 2 and 3 5

1 and 6 7

-2 and -3 -5

-1 and -6 -7

- 2 - 3

Factors of 6 Sum of Factors

Using zero property

x – 2 = 0 or x – 3 = 0

x = 2 or x = 3

Solving Q. E. by Factoring

Example: Factor and use the zero property to solve the Q.E.

3. x² − 2x − 3 = 0

= 0( x ) ( x ) 1 and -3 -2

-1 and 3 2+ 1 - 3

Factors of -3 Sum of Factors

Using zero property

x + 1 = 0 or x – 3 = 0

x = -1 or x = 3

Solving Q. E. by Factoring

Example: Box Method Factoring and the zero property to solve the Q.E.

F of -18 Sum

-9, 2 -7

9,-2 7

-6,3 -3

6,-3 3

1,-18 -17

-1,18 17

1. 3x2 + 7x – 6 = 0

1st: (3)(-6) = -18

2nd:3rd: Box Method

3x2

-6

9x

-2x

x 3

3x

-2

(x + 3)(3x – 2) = 0

GCF

x + 3 = 03x – 2 = 0+2 +2

3x = 23 3

x 23

=

x =-3

Friday, January 31, 2020

Solving Q. E. by Factoring

F of -168 Sum

-21, 8 -13

21,-8 13

-12,14 2

12,-14 -2

28,-6 22

-28,6 17

2. 8x2 + 22x – 21 = 0

1st: (8)(-21) = -168

2nd:3rd: Box Method

8x2

-21

28x

-6x

2x 7

4x

-3

(2x + 7) (4x – 3) = 0

GCF

Example: Box Method Factoring and the zero property to solve the Q.E.

4x – 3 = 0+3 +3

4x = 34 4

x 34

=

2x + 7 = 0- 7 -7

2x = -72 2

x -72

=

Solving Quadratic Equation

Example: Solve by factoring

1.) 6x2 + 7x – 5 = 0

2.) x2 – 6x = 27

(2x - 1)(3x + 5) = 0

(2x - 1) = 0 Or (3x + 5) = 0

x2 – 6x – 27 = 0

(x + 3)(x – 9) = 0

(x + 3) = 0 Or (x – 9) = 0

x = -3 Or x = 9

x = 1/2 Or x = -5/3

Solving Quadratic Equation

Solving by Completing the Square

Recall: Perfect Square Trinomials

Examples

x2 + 6x + 9

x2 - 10x + 25

x2 + 12x + 36

= (x + 3)(x + 3)

= (x + 3)2

= (x + 6)(x + 6)

= (x – 5)(x – 5)

= (x – 5)2

= (x + 6)2

Friday, January 31, 2020

Creating a Perfect Square Trinomial

In the following perfect square trinomial, the constant term is missing. X2 + 14x + ____

Find the constant term by squaring half the coefficient of the linear term (the number beside x).

(14/2)2

X2 + 14x + 49

Creating a Perfect Square Trinomial

Create a perfect square trinomial and factor

x2 + 20x + ___

x2 - 4x + ___

x2 + 5x + ___

100

4

25/4

= (x + 10)2

= (x – 2)2

= (x + 5/2)2

Solving Quadratic Equation

Solving by Completing the Square

Friday, January 31, 2020

Solving Quadratic Equation

Solve by Completing the Square Example 1

Step 1: Move quadratic term, and linear term to left side and the constant term to right side of the equation

x2 + 8x – 20 = 0

+ 20 +20

x2 + 8x = 20Step 2: Find the number that completes the square on the left side and add to both sides.of equation

x2 + 8x = 20+ 16 +16

Step 3: Factor the left side of the equation and simplify the right side of the equation

(x + 4)2 = 36

Solve by Completing the Square

2( 4) 36x

( 4) 6x

Solving Quadratic Equation

Step 4: Solve by taking the square root

4 6

4 6 an

d 4 6

10 and 2 x=

x

x x

x

Solve by Completing the Square Example 2

22 7 12 0x x

22 7 12x x

Solving Quadratic Equation

Step 1: Move quadratic term, and linear term to left side and the constant term to right side of the equation

Step 2:

Find the term that completes the square on the left side of the equation. Add that term to both sides.

2

2

2

2 7

2

2 2 2

7 12

7

2

=-12 +

6

x x

x x

xx

21 7 7 49

( ) then square it, 2 62 4 4 1

7

2 49 49

16 1

76

2 6x x

Solving Quadratic Equation

The quadratic coefficient must be equal to 1 before you complete the square, so you must divide all terms by the quadratic coefficient first.

Step 3:

Factor the perfect square trinomial on the left side of the equation. Simplify the right side of the equation.

2

2

2

76

2

7 96 49

4 16 16

7 47

4

49 49

16 1

16

6x x

x

x

Solving Quadratic Equation

27 47( )

4 16x

7 47( )

4 4

7 47

4 4

7 47

4

x

ix

ix

Solving Quadratic Equation

Step 4: Solve by taking the square root

2

2

2

2

2

1. 2 63 0

2. 8 84 0

3. 5 24 0

4. 7 13 0

5. 3 5 6 0

x x

x x

x x

x x

x x

Try the following examples. Do your work on your paper and then check your answers.

1. 9,7

2.(6, 14)

3. 3,8

7 34.

2

5 475.

6

i

i

Solving Quadratic Equation

Solve by Completing the Square

Solving Quadratic Equation

Solving by Quadratic Formula

ax2 + bx + c = 0

ax2 + bx = -c

a

cx

a

bx2

2

2

2

22

a4

b

a

c

a4

b x

a

bx

The quadratic formula is derived by completing the square using the standard quadratic equation:

2

22

a4

ac4b

a2

bx

Solving Quadratic Equation

Solving by Quadratic Formula

a2

ac4bbx

2

The Quadratic Formula is

The standard quadratic equation form is ax2 + bx + c = 0 where a, b, and c are numbers with a 0.

All quadratic equation can be solved by using the quadratic formula

Friday, January 31, 2020

Solving Quadratic Equation

Solve by quadratic formula

1.) 2x2 + 4x = 5 2x2 + 4x – 5 = 0

a2

ac4bbx

2

a=2 ,b=4, and c=-5

)2(2

)5)(2(4)4(4 2 x

4

40164x

4

564x

4

1424x

2

142x

Solving Quadratic Equation

Solve by quadratic formula

2.) x2 + 13 = 4x x2 – 4x + 13 = 0

a2

ac4bbx

2

a=1 ,b=-4, and c=13

)1(2

)13)(1(4)4()4(x

2

2

52164x

2

364x

2

i64x

i32x

Changing equation in Standard

Change the equation in standard quadratic form and identify a,b, and c then solve using Quadratic Formula

1.) 2x2 + 8 = 0

4.) (x – 3)2 + 8 = 44

2.) 3x2 = 9x

3.) x2 + 4x = 5

a=2 ,b=0, and c=8

3x2 – 9x = 0

2x2 + 8 = 0

a=3 ,b=-9, and c=0

x2 + 4x – 5 = 0

a=1 ,b=4 and c=-5

x2 – 6x – 27 = 0

a=1 ,b=-6, and c=-27

ax2 + bx + c = 0Standard form

Solving Quadratic Equation

Solve by quadratic formula

1.) x2 – 3x – 2 = 0

2.) 2x2 + 13 = 8x

3.) 3x2 – 3 = 4x

Find Quadratic Equation

Finding a Quadratic Function or Equation given its Roots, Zeros or Solution.

Example:

1. Find a Quadratic function/ equation whose zeros are x = 1 and x = -3

Solution:

f(x) =

(x – 1) (x + 3)

f(x) = x2 + 3x - 1x - 3

f(x) = x2 + 2x - 3 or x2 + 2x – 3 = 0

Find Quadratic Equation

Finding a Quadratic Function or Equation given its Roots, Zeros or Solution.

Try it yourself:

2. Find a Quadratic function/ equation whose zeros are x = 2 and x = -1

Solution:

f(x) =

(x – 2) (x + 1)

f(x) = x2 + 1x - 2x - 2

f(x) = x2 - 1x - 2 or x2 - 1x – 2 = 0