Forms of a Quadratic Equation

-4x² + 4x =1 x(x - 2) = 0

-4x² + 4x -1 = 0 x² - 2x = 0

(x – 4)(x + 4) = 9

x² - 16 = 9x² - 25 = 0

SOLVING EQUATIONS USING THE ZERO PRODUCT RULE

Definition of a Quadratic Equation in One VariableIf a, b, and c are real numbers such that a ≠ 0, then a quadratic equation is an equation that can be written in the form

ax² + bx + c = 0

Zero Product Rule

If ab = 0, then a = 0 or b = 0

2 12 0x x Factor

( 4)( 3) 0x x

( 4) 0x

Apply the zero product rule

3 0x

Set each factor equal to zero

Solve each equation for x4x 3x

Steps for Solving a Quadratic Equation by Factoring

1. Write the equation in the form: ax² + bx + c = 0

2. Factor the equation completely.3. Apply the zero product rule, that is, set

each factor equal to zero, and solve the resulting equations.

Note: The solution(s) found in Step 3 may be checked by substitution into the original equation.

Solve Quadratic Equation22 9 5x x Write the equation in the

formax² + bx +c = 0

22 9 5 0x x Factor the polynomial completely

(2 1)( 5) 0x x Set each factor equal to zero

2 1 0x 5 0x Solve each equation

1

2x 5x Solutions

Solve Quadratic Equation22 9 5x x Write the equation in the

formax² + bx +c = 0

22 9 5 0x x Factor the polynomial completely

(2 1)( 5) 0x x Set each factor equal to zero

2 1 0x 5 0x Solve each equation

1

2x 5x Solutions

Check: x = -½22 9 5x x

Check: x = 5 22 9 5x x

22(5) 9(5) 5

1 92( ) 54 2

1 95

2 2

105

2

21 12( ) 9( ) 52 2

2(25) 45 5

50 45 5

Solve Quadratic Equation

5 (5 2) 10 9x x x Write the equation in the form

ax² + bx +c = 02

2

2

25 10 10 9

25 10 10 9 0

25 9 0

x x x

x x x

x

Factor the polynomial completely

(5 3)(5 3) 0x x Set each factor equal to zero

5 3 0x 5 3 0x Solve each equation

3

5x 3

5x

Solutions

Solve Higher Quadratic Equation3 25 9 45 0w w w This is a higher degree

polynomial equation.

3( 5 ) ( 9 45) 0w w w The equation is already set equal to zero. Because there are four terms, try factoring by grouping.

Solve each equation

Solutions

2

2

( 5) 9( 5) 0

( 5)( 9) 0

( 5)( 3)( 3) 0

w w w

w w

w w w

5 0w 3 0w 3 0w

5w 3w 3w

Translating to Quadratic Equation

( 1) ( 1) 48x x x

Factor the polynomial completely

( 7)( 7) 0x x Set each factor equal to zero

7 0x 7 0x Solve each equation7x 7x

Solutions

The product of two consecutive integers is 48 more than the larger integer. Find the integers.

Let x represent the first (smaller) integer.Then x + 1 represents the second (larger) integer

Simplify

2

2

49

49 0

x x x

x

hypotenus

e

leg

leg

In a right triangle, the shorter sides are called legs and the longest side (which is the one opposite the right angle) is called the hypotenuse

The Pythagorean Theorem relates the lengths of the sides.

a

b

c

c2 = a2 + b2

If one leg of a right triangle measures 3 inches and one leg measures 4 inches, how long is the hypotenuse?

c2 = (3)2 + (4)2

c2 = 9 + 16 = 25

c = 5

Applying the Pythagorean Theorem

a

c =10

b =6

Apply the Pythagorean theorem.

a² + b² = c²

Substitute b= 6 and c = 10a² + 6² = 10²

a² + 36 = 100Simplify

a² + 36 -100 = 100 - 100a² - 64 = 0 Factor

(a + 8)(a - 8) = 0Set each factor equal to zero

a + 8 =0 or a – 8 = 0

Because x represents the length of a side of a triangle, reject the negative solution.

a = -8 a = 8