1. Write 15x2 + 6x = 14x2 – 12 in standard form.

2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5.

ANSWER

ANSWER –24

x2 + 6x +12 = 0

Classwork Answers…4.8 (13-27 odd)

4.6 (multiples of 3)

a

acbbx

2

42

When solving quadratic equations, we’re looking for the x values where the graph crosses the x axis

One of the methods we use to solve quadratic equations is called the Quadratic Formula

a

acbbx

2

42

Using the a, b, and c from ax2 + bx + c = 0

Must be equal to zero

EXAMPLE 1 Solve an equation with two real solutions

Solve x2 + 3x = 2.x2 + 3x = 2 Write original equation.

x2 + 3x – 2 = 0 Write in standard form.

x =– b + b2 – 4ac2a

Quadratic formula

x =– 3 + 32 – 4(1)(–2) 2(1)

a = 1, b = 3, c = –2

Simplify.x = – 3 + 172

56.02

173

x or 56.3

2

173

x

EXAMPLE 2 Solve an equation with one real solutions

Solve 25x2 – 18x = 12x – 9.25x2 – 18x = 12x – 9. Write original equation.

Write in standard form.

x =30 + (–30)2– 4(25)(9)2(25)

a = 25, b = –30, c = 9

Simplify.

25x2 – 30x + 9 = 0.

x =30 + 0

50

x = 35 Simplify.

35The solution is

ANSWER

EXAMPLE 3 Solve an equation with imaginary solutions

Solve –x2 + 4x = 5.–x2 + 4x = 5 Write original equation.

Write in standard form.

x =–4 + 42 – 4(–1)(–5)2(–1)

a = –1, b = 4, c = –5

Simplify.

–x2 + 4x – 5 = 0.

x =–4 + –4

–2–4 + 2i

x = –2

Simplify.

Rewrite using the imaginary unit i.

x = 2 + i

The solution is 2 + i and 2 – i.

ANSWER

GUIDED PRACTICE for Examples 1, 2, and 3

Use the quadratic formula to solve the equation.

x2 = 6x – 41.

x2 – 6x + 4 = 0

a = 1 b = -6 c = 4

x =– b + b2 – 4ac2a

)1(2

)4)(1(4)6(6 2 x

2

206 x

2

526 x

53x

2

)53(2 x

GUIDED PRACTICE for Examples 1, 2, and 3

Use the quadratic formula to solve the equation.

4x2 – 10x = 2x – 92.

4x2 – 12x + 9 = 0

a = 4 b = -12 c = 9

x =– b + b2 – 4ac2a

)4(2

)9)(4(4)12(12 2 x

8

012 x

8

12x

2

3x

EXAMPLE 4 Use the discriminant

EquationDiscriminant

Solution(s)

If the quadratic equation is in the standard form ax2 + bx + c = 0

b2 – 4ac

a. x2 – 8x + 17 = 0 ( –8)2 – 4(1)(17) = –4 Two imaginary

b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real

b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 4 Two real

The discriminant can be found using b2 – 4ac

If b2 – 4ac < 0 There are Two Imaginary solutions

If b2 – 4ac = 0 There is One Real solution

If b2 – 4ac > 0 There are Two Real solutions

GUIDED PRACTICE for Example 4

Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.

4. 2x2 + 4x – 4 = 0 5. 3x2 + 12x + 12 = 0

6. 8x2 = 9x – 11 7. 4x2 + 3x + 12 = 3 – 3x

(4)2 – 4(2)(-4) = 48

b2 – 4ac

Two Real solutions

(12)2 – 4(3)(12) = 0

b2 – 4ac

One Real solution

(-9)2 – 4(8)(11) = -271

b2 – 4ac

Two Imaginary solutions

(6)2 – 4(4)(9) = -108

b2 – 4ac

Two Imaginary solutions

8x2 – 9x + 11 = 0 4x2 + 6x + 9 = 0

Classwork Assignment:WS 4.8 (1-27 odd)