Use the Quadratic Formula to solve 3x2 + 23x + 40 = 0. Lesson 5-8 Additional Examples 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 ± 232 – 4(3)(40) 2(3) Substitute. –b ± b2 – 4ac 2a –23 ± 7 6 = –23 ± 49 –23 ± 529 – 480 Simplify. = – or – 30 6 16 = –5 or – 8 3

(continued) – + 40 0 Check: 3x2 + 23x + 40 = 0 3x2 + 23x + 40 = 0 The Quadratic Formula Lesson 5-8 Additional Examples (continued) – + 40 0 Check: 3x2 + 23x + 40 = 0 3x2 + 23x + 40 = 0 3(–5)2 + 23(–5) + 40 0 75 – 115 + 40 0 0 = 0 3 – + 23 – + 40 0 8 3 64 184 2

Find the values of a, b, and c. The Quadratic Formula Lesson 5-8 Additional Examples 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

Find the values of a, b, and c. The Quadratic Formula Lesson 5-8 Additional Examples 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.

Is the answer reasonable? Since is a negative number, and a The Quadratic Formula Lesson 5-8 Additional Examples (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 1.56, that answer is reasonable. 17 – 1

Determine the type and number of solutions of x2 + 5x + 10 = 0. The Quadratic Formula Lesson 5-8 Additional Examples 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.

Write the equation in standard form. The Quadratic Formula Lesson 5-8 Additional Examples 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 2 + 25t + 6. If the ball makes it to where the wall is, will it go over the wall or hit the wall? h = –16t 2 + 25t + 6 17 = –16t 2 + 25t + 6 Substitute 17 for h. 0 = –16t 2 + 25t – 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.