1. Section 5.8
Quadratic Formula

2. Write each quadratic equation in standard form.
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
(For help, go to Lessons 1-2 and 5-1.)
Write each quadratic equation in standard form.
1. y = 8 – 10x2 2. y = (x + 2)2 – 1
3. a = 1, b = 6, c = 3 4. a = –5, b = 2, c = 4
y = –10x2 + 8
y = x2 + 4x + 3
Evaluate the expression b2 – 4ac for the given values of a, b, and c.
62 – 4(1)(3) = 36 – 12 = 24
22 – 4(–5)(4) = 4 – (–80) = 84
Check Skills You’ll Need
5-8

3. Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0.
ALGEBRA 2 LESSON 5-8
Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0.
3x2 + 23x + 40 = 0
a = 3, b = 23, c = 40
Write in standard form.
Find values of a, b, and c.
x =
Write the Quadratic Formula. =
–23 ± – 4(3)(40)
2(3)
Substitute.
–b ± b2 – 4ac 2a
–23 ± 7 6 =
–23 ± 49
–23 ± – 480
Simplify.
= – or – 30 6 16
= –5 or – 8 3
5-8

4. Find the values of a, b, and c.
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Solve 3x2 + 2x = –4.
3x2 + 2x + 4 = 0
Write in standard form.
a = 3, b = 2, c = 4
Find the values of a, b, and c.
–(2) ± (2)2 – 4(3)(4)
2(3)
y =
Substitute.
–2 ± 4 – 48 6
–2 ± –44 =
–2 ± 2i 11
Simplify.
i 11 3
= – ± 1
Quick Check
5-8

5. Find the values of a, b, and c.
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
The longer leg of a right triangle is 1 unit longer than the shorter leg. The hypotenuse is 3 units long. What is the length of the shorter leg?
x2 + (x + 1)2 = 32
x2 + x2 + 2x + 1 = 9
2x2 + 2x – 8 = 0
Write in standard form.
–b ± b – 4ac 2a 2
a = 2, b = 2, c = –8
x =
Find the values of a, b, and c.
Use the Quadratic Formula. =
–(2) ± (2)2 – 4(2)(–8)
2(2)
Substitute for a, b, and c.
–1 ± 17 2 =
Simplify.
17 – 1 2
The length of the shorter leg is units.
5-8

6. Is the answer reasonable? Since is a negative number, and a
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
(continued)
– 1 – 17 2
Is the answer reasonable? Since is a negative number, and a
length cannot be negative, that answer is not reasonable.
Since , that answer is reasonable.
17 – 1
Quick Check
5-8

7. Determine the type and number of solutions of x2 + 5x + 10 = 0.
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Determine the type and number of solutions of
x2 + 5x + 10 = 0.
a = 1, b = 5, c = 10 Find the values of a, b, and c.
b2 – 4ac = (5)2 – 4(1)(10) Evaluate the discriminant.
= 25 – 40 Simplify.
= –15
Since the discriminant is negative, x2 + 5x + 10 = 0 has two imaginary solutions.
Quick Check
5-8

8. Write the equation in standard form.
The Quadratic Formula
ALGEBRA 2 LESSON 5-8
Quick Check
A player throws a ball up and toward a wall that is 17 ft high. The height h in feet of the ball t seconds after it leaves the player’s hand is modeled by h = –16t t + 6. If the ball makes it to where the wall is, will it go over the wall or hit the wall?
h = –16t t + 6
17 = –16t t + 6
Substitute 17 for h.
0 = –16t t – 11
Write the equation in standard form.
a = –16, b = 25, c = –11
Find the values of a, b, and c.
b2 – 4ac = (25)2 – 4(–16)(–11)
Evaluate the discriminant.
= 625 – 704
= –79
Simplify.
Since the discriminant is negative, the equation 17 = –16t2 + 25t + 6 has no real solution. The ball will hit the wall.
5-8

9. Use the Quadratic Formula to solve each equation.
ALGEBRA 2 LESSON 5-8
Use the Quadratic Formula to solve each equation.
1. 6x2 + 19x + 8 = 0 2. x2 – 2x – 11 = 0
3. x2 – 2x – 15 = x2 – 7x + 5 = 0
5. 2x2 – 5x + 7 = 0 6. –3x2 – 14x – 8 = 0
7. 4x2 – 5x + 10 = 7x + 1
– , – 8 3 1 2
1 ±
7 ± i 6
–3, 5
Find the value of the discriminant for each quadratic equation. Tell how many different solutions each equation has and whether the solutions are real or imaginary.
–31; two, imaginary
100; two, real
0; one, real
5-8