1. Lesson 9-8 Warm-Up

2. “Using the Discriminant” (9-8)
What is the “discriminant”?
The discriminant is the expression under the radical in a quadratic formula which you can use to determine if the quadratic equation has one, two, or no solutions (Note: a ≠ 0) If the discriminant is positive (b2 - 4ac  0), there are two solutions or x-intercepts. If the discriminant is zero (b2 - 4ac = 0), there is one solution or x-intercept. If the discriminant negative (b2 - 4ac  0), there are no solutions or x-intercepts. Examples: y = x2 - 6x + 3 y = x2 - 6x + 9 y = x2 - 6x + 12 a = 1, b = -6, c = 3 a = 1, b = -6, c = 9 a = 1, b = -6, c = 12 b2 - 4ac discriminant b2 - 4ac discriminant b2 - 4ac (-6)2 - 4(1)(3) (-6)2 - 4(1)(9) (-6)2 - 4(1)(12) 36 – 12 = 24 (Positive) 36 – 36 = 0 (Zero) 36 – 48 = -12 (Negative) There are 2 x-intercepts. There is 1 x-intercept. There are no x-intercepts.

3. Determine whether the graph of y = 3x2 – 13x – 10
Using the Discriminant
LESSON 9-8
Additional Examples
Determine whether the graph of y = 3x2 – 13x – 10
intersects the x-axis in zero, one, or two points.
Method 1: Use the discriminant.
b2 – 4ac = (– 13)2 – (4)(3)(– 10)
Evaluate the discriminant.
Substitute for a, b, and c.
= Use the order of operations.
= Simplify.
The function has two zeros, so the graph intersects the x-axis in two
points.

4. Set the expression equal to zero.
Using the Discriminant
LESSON 9-8
Additional Examples
(continued)
Method 2 Factor.
3x2 – 13x – 10 = 0
Set the expression equal to zero.
(3x + 2)(x – 5) = 0
Factor.
3x + 2 = or x – 5 = 0
Use the Zero-Product Property.
x = – or x = 0 2 3
Solve for x.
The function has two zeros, so the graph intersects the x-axis in two points.

5. Find the number of solutions of x2 = –3x – 7.
Using the Discriminant
LESSON 9-8
Additional Examples
Find the number of solutions of x2 = –3x – 7.
x2 + 3x + 7 = 0 Write in standard form.
b2 – 4ac = 32 – 4(1)(7) Evaluate the discriminant. Substitute for a, b, and c.
= 9 – 28 Use the order of operations.
= –19 Simplify.
Since –19 < 0, the equation has no solution.

6. h = –16t2 + vt + c Use the vertical motion formula.
Using the Discriminant
LESSON 9-8
Additional Examples
A football is punted from a starting height of 3 ft with an initial upward velocity of 40 ft/s. Will the football ever reach a height of 30 ft?
h = –16t2 + vt + c Use the vertical motion formula.
30 = –16t2 + 40t Substitute 30 for h, 40 for v, and 3 for c.
0 = –16t2 + 40t – Write in standard form.
b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant.
= 1600 – Use the order of operations.
= – Simplify.
The discriminant is negative. The football will never reach a height of 30 ft.

7. Find the number of solutions for each equation. 1. 3x2 – 4x = 7
Using the Discriminant
LESSON 9-8
Lesson Quiz
Find the number of solutions for each equation.
1. 3x2 – 4x = 7
2. 4x2 = 4x – 1
3. –3x2 + 2x – 12 = 0
4. A ball is thrown from a starting height of 4 ft with an initial upward velocity of 30 ft/s. Is it possible for the ball to reach a height of 18 ft?
two
one
none
yes