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§9.6 The Quadratic Formula & the Discriminant

§9.6 The Quadratic Formula & the Discriminant. Warm-Up. Find the value of c to complete the square for each expression. 1. x 2 + 6 x + c 2. x 2 + 7 x + c 3. x 2 – 9 x + c. Solve each equation by completing the square. 4. x 2 – 10 x + 24 = 0 5. x 2 + 16 x – 36 = 0

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§9.6 The Quadratic Formula & the Discriminant

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  1. §9.6 The Quadratic Formula & the Discriminant

  2. Warm-Up Find the value of c to complete the square for each expression. 1.x2 + 6x + c2.x2 + 7x + c3.x2 – 9x + c Solve each equation by completing the square. 4.x2 – 10x + 24 = 0 5.x2 + 16x – 36 = 0 6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0

  3. 6 2 –10 2 –9 2 7 2 16 2 1.x2 + 6x + c; c = 2 = 32 = 9 2.x2 + 7x + c; c = 2 = 3.x2 – 9x + c; c = 2 = 4.x2 – 10x + 24 = 0; 5. x2 + 16x – 36 = 0; c = 2 = (–5)2 = 25 c = 2 = 82 = 64 x2 – 10x = –24 x2 + 16x = 36 x2 – 10x + 25 = –24 + 25 x2 + 16x + 64 = 36 + 64 (x – 5)2 = 1 (x + 8)2 = 100 (x – 5) = ±1 (x + 8) = ±10 x – 5 = 1   or x – 5 = –1 x + 8 = 10  or x + 8 = –10 x = 6   or   x = 4 x = 2   or  x = –18 Solutions 49 4 81 4

  4. 4 2 –1 2 Solutions 6. 3x2 + 12x – 15 = 0 7. 2x2 – 2x – 112 = 0 3(x2 + 4x – 5) = 0 2(x2 – x – 56) = 0 x2 + 4x – 5 = 0; x2 – x – 56 = 0; c = 2 = 22 = 4 c = 2 = x2 + 4x = 5 x2 – x = 56 x2 + 4x + 4 = 5 + 4 x2 – x + = 56 + (x + 2)2 = 9 (x – )2 = (x + 2) = ±3 x – = ± x + 2 = 3   or x + 2 = –3 x – = or x – = x = 1   or   x = –5 x = 8 or x = –7 1 4 1 4 1 4 1 2 225 4 1 2 15 2 1 2 15 2 1 2 –15 2

  5. Formula Formula • The Quadratic Formula • If ax2+ bx + c = 0, and a≠0, then: • x = – b ± √ b2 – 4ac 2a

  6. x2 + 3x + 2 = 0 Add 3x to each side and write in standard form. x = Use the quadratic formula. –b± b2– 4ac 2a Substitute 1 for a, 3 for b, and 2 for c. x = Simplify. x = Write two solutions. –3 ± 1 2 –3 ± (–3)2– 4(1)(2) 2(1) x = or x = –3 + 1 2 –3 – 1 2 x = –1 or x = –2 Simplify. Example 1: Using the Quadratic Formula Solve x2 + 2 = –3x using the quadratic formula.

  7. Check: for x = –1 for x = –2 (–1)2 + 3(–1) + 2 0 (–2)2 + 3(–2) + 2 0 1 – 3 + 2 0 4 – 6 + 2 0 0 = 0 0 = 0 Example 1: Using the Quadratic Formula

  8. x = Use the quadratic formula. –b± b2– 4ac 2a Substitute 3 for a, 4 for b, and –8 for c. x = –4 ± 112 6 –4 – 112 6 –4 + 112 6 x = or x = Write two solutions. x = x or x Use a calculator. –4 ± 42– 4(3)(–8) 2(3) –4 + 10.583005244 6 –4 – 10.583005244 6 Round to the nearest hundredth. x –2.43 x 1.10 or Example 2: Finding Approximate Solutions Solve 3x2 + 4x– 8 = 0. Round the solutions to the nearest hundredth.

  9. Example 3: Choosing an Appropriate Method A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second. Step 1: Use the vertical motion formula. h = –16t2 + vt + c 0 = –16t2 + 15t + 2Substitute 0 for h, 15 for v, and 2 for c.

  10. –b± b2– 4ac 2a Step 2: Use the quadratic formula. x = • ExampleΩ: Real-World Problem Solving A child throws a ball upward with an initial upward velocity of 15 ft/s from a height of 2 ft. If no one catches the ball, after how many seconds will it land? Use the vertical motion formula h = –16t2 + vt + c, where h = 0, v = velocity, c = starting height, and t = time to land. Round to the nearest hundredth of a second.

  11. Substitute –16 for a, 15 for b, 2 for c, and t for x. Simplify. t = –15 ± 225 + 128 –32 t = t = –15 + 18.79 –32 –15 – 18.79 –32 Write two solutions. –15 ± 353 –32 t = t = or Simplify. Use the positive answer because it is the only reasonable answer in this situation. or t –0.12 –15 ± 152– 4(–16)(2) 2(–16) t 1.06 ExampleΩ: Real-World Problem Solving The ball will land in about 1.06 seconds.

  12. Example 3: Choosing an Appropriate Method Which method(s) would you choose to solve each equation? Justify your reasoning. Quadratic formula; the equation cannot be factored easily. a. 5x2 + 8x – 14 = 0 b. 25x2 – 169 = 0 Square roots; there is no x term. Factoring; the equation is easily factorable. c.x2 – 2x – 3 = 0 Quadratic formula, completing the square, or graphing; the x2 term is 1, but the equation is not factorable. d.x2 – 5x + 3 = 0 e. 16x2 – 96x + 135 = 0 Quadratic formula; the equation cannot be factored easily and the numbers are large.

  13. Property Property • Property of the Discriminant • For the quadratic equation ax2 + bx + c = 0, where a≠0, you can use the value of the discriminant to determine the number of solutions. The discriminant is defined as: b2 – 4ac. • If b2 – 4ac > 0, there are two solutions. • If b2 – 4ac = 0, there are one solutions. • If b2 – 4ac < 0, there are no solutions.

  14. Example 4: Choosing an Appropriate Method Find the number of solutions of x2 = –3x – 7 using the discriminant. 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.

  15. Assignment: Pg. 586-587 7-32 Left

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