Lesson 9-8 Warm-Up

“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.

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. = 169 + 120 Use the order of operations. = 289 Simplify. The function has two zeros, so the graph intersects the x-axis in two points.

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 = 0 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.

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.

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 + 3 Substitute 30 for h, 40 for v, and 3 for c. 0 = –16t2 + 40t – 27 Write in standard form. b2 – 4ac = (40)2 – 4 (–16)(–27) Evaluate the discriminant. = 1600 – 1728 Use the order of operations. = –128 Simplify. The discriminant is negative. The football will never reach a height of 30 ft.

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