Warm-Up Exercises Write in vertex form. 1. ANSWER ( )2)2 x 3 + y 3 = + 2. Evaluate when,, and. 3 = a b 2b 2 4ac – 6 = b – 5 = c ANSWER 24 – ANSWER 25 ( )2)2 x 3 – = A student is solving an equation by completing the square. Write the step in the solution that appears just before “ ” – 5 ( ) x 3 – = y = x 2x 2 6x6x12 + +

Example 1 Solve an Equation with Two Real Solutions Solve. = x 2x 2 +3x3x5 – 0 SOLUTION = x 2x 2 +3x3x5 – 0 Write original equation. = x 2a2a b – + – 4ac4acb 2b 2 – Quadratic formula = x 3 – + – – () 1 () 5 – 2 () 1 Substitute values in the quadratic formula: a 1, b 3, and c 5. – = == = x 3 – + – 29 2 Simplify.

Example 1 Solve an Equation with Two Real Solutions ANSWER The solutions are and. 3 – ≈ –– 29 2 ≈ 4.19 –

Checkpoint Use the quadratic formula to solve the equation. Solve an Equation with Two Real Solutions 1. = x 2x 2 +2x2x3 – 0 2. = 2x 22x 2 +4x4x1 – 0 ANSWER 2 – ≈ 0.22, 2 – 6 2 ≈ 2.22 – – ANSWER 3,3, – 1

Checkpoint Use the quadratic formula to solve the equation. Solve an Equation with Two Real Solutions 3. = 3x 23x 2 2x2x6 – 0 – ANSWER ≈ ≈ 1.12 – –

Example 2 Solve an Equation with One Real Solution Solve x 2x 2 6x6x – = – 9.9. SOLUTION x 2x 2 6x6x – = – 9 Write original equation. x 2x 2 6x6x 9 – + = 0 Write equation in standard form. 2a2a x = – b + – b 2b 2 – 4ac4ac Quadratic formula

6 2 Example 2 Solve an Equation with One Real Solution Simplify. x = + – x = 3 ANSWER The solution is 3. x = + – ( – 6 ( – ( – ( – 1 9 ( (( ( 4 1 ( ( 2 Substitute values in the quadratic formula: a 1, b 6, and c 9. = == –

Example 3 Solve an Equation with Imaginary Solutions Solve x 2x 2 2x2x = SOLUTION Write original equation. x 2x 2 2x2x = x = + – – 2 – 1 2 ( (( ( 4 1 ( ( 2 2 Substitute values in the quadratic formula: a 1, b 2, and c 2. = == 2a2a x = – b + – b 2b 2 – 4ac4ac Quadratic formula x = + – – 2 – 4 2 Simplify.

Example 3 Solve an Equation with Imaginary Solutions Simplify and rewrite using the imaginary unit i. x = + – – 2 2 2i2i + – 1 i ANSWER The solutions are and. – 1 i – Simplify. x = + – – 1 i

Checkpoint Use the quadratic formula to solve the equation. Use the Quadratic Formula 4. x 2x 2 2x2x 4 – = 0 – 5. –x 2x 2 1 – = – 3x3x ANSWER 15 + ~ ~ 3.24, 15 ~ ~ –– 1.24 ANSWER ~ ~ 0.38 – ~ ~ 2.62,

Checkpoint Use the quadratic formula to solve the equation. Use the Quadratic Formula 6. 2x 22x = 5x5x – ANSWER 2 7 – 1,1, 7. x 2x 2 4x4x = ANSWER + – – 2 i x 2x 2 2x2x3 + 0 = –– 8. ANSWER + – 1 i 2

Checkpoint Use the quadratic formula to solve the equation. Use the Quadratic Formula 9. 2x 22x 2 x = 4 – + ANSWER + – 1 i 4 31

Example 4 Use the Discriminant Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. x 2x 2 6x6x8 + 0 = – a. x 2x 2 6x6x9 + 0 = – b. x 2x 2 6x6x = – c. 0 SOLUTION EQUATIONDISCRIMINANTTYPE OF SOLUTION(S) ax 2 bxc + = + b 2b 2 4ac – = b 2b 2 – – + b – 2a2a x x 2x 2 6x6x8 + 0 = – a ( ( – 4 – ( ( 1 ( ( 8 = Two real

Example 4 Use the Discriminant x 2x 2 6x6x9 + 0 = – b.6 2 ( ( – 4 – ( ( 1 ( ( 9 0 = One real x 2x 2 6x6x = – c.6 2 ( ( – 4 – ( ( 1 ( ( 10 = Two imaginary 4 –

Checkpoint Find the discriminant of the quadratic equation and give the number and type of solutions of the equation. Use the Discriminant 10. ANSWER – 8, 2 imaginary solutions x 2x 2 2x2x = ANSWER 25, 2 real solutions 11. x 2x 2 3x3x = – 12. x 2x 2 4x4x4 + 0 = – ANSWER 0, 1 real solutions

Example 5 Model Vertical Motion A baton twirler tosses a baton into the air. The baton leaves her hand when it is 6 feet above the ground. The initial vertical velocity of the baton is 30 feet per second. The baton twirler catches the baton when it falls back to a height of 5 feet. Baton Twirling a.Write an equation that gives the height (in feet) of the baton as a function of time (in seconds). b. For how many seconds is the baton in the air?

Example 5 Model Vertical Motion SOLUTION a.The baton is thrown, so use the model with and – = 16t 2 h + v0tv0t + h0h0 = 30v0v0 = 6.6. h0h0 Substitute 30 for v 0 and 6 for h 0. – = 16t 2 h + 30t + 6 Write vertical motion model for a thrown object. – = 16t 2 h + v0tv0t + h0h0

Example 5 Model Vertical Motion Substitute 5 for h. – = 16t t + 6 Write in standard form. – = 16t t + 1 t Use a calculator. ~ ~ 1.9 t ~ ~ 0.03 or – Write model from part (a). – = 16t 2 h + 30t + 6 t = + – ( – 3030 ( – 30 2 – 1 ( ( 4 ( – 1616 ( Substitute values in the quadratic formula: a 16, b 30, and c 1. = = = – 2 ( – 1616 ( b. To find the number of seconds the baton is in the air, find the number of seconds that have passed when the height of the baton is 5 feet. Find the value of t when h 5. =

Example 5 Model Vertical Motion CHECKYou can check your results by graphing and finding when y 0. Because the graph crosses the x -axis at about x 1.9, your answer is correct. – = 16x 2 y + 30x + 1 = = ANSWER The solution doesn’t make sense, because time cannot be negative. The baton is in the air for about 1.9 seconds –