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Section 5.8 Quadratic Formula

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5-8 1.y = 8 – 10x 2 2.y = (x + 2) 2 – 1 3.a = 1, b = 6, c = 34.a = –5, b = 2, c = 4 (For help, go to Lessons 1-2 and 5-1.) ALGEBRA 2 LESSON 5-8 The Quadratic Formula Write each quadratic equation in standard form. Evaluate the expression b 2 – 4ac for the given values of a, b, and c. Check Skills You’ll Need y = –10x 2 + 8 y = x 2 + 4x + 3 6 2 – 4(1)(3) = 36 – 12 = 24 2 2 – 4(–5)(4) = 4 – (–80) = 84

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Use the Quadratic Formula to solve 3x 2 + 23x + 40 = 0. ALGEBRA 2 LESSON 5-8 The Quadratic Formula = –or – 30 6 16 6 = –5 or – 8383 3x 2 + 23x + 40 = 0 a = 3, b = 23, c = 40 Write in standard form. Find values of a, b, and c. –23 ± 7 6 = –23 ± 49 6 = –23 ± 529 – 480 6 Simplify. = x = Write the Quadratic Formula. = –23 ± 23 2 – 4(3)(40) 2(3) Substitute. –b ± b 2 – 4ac 2a 5-8

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Solve 3x 2 + 2x = –4. ALGEBRA 2 LESSON 5-8 The Quadratic Formula i 11 3 = – ± 1313 3x 2 + 2x + 4 = 0Write in standard form. a = 3, b = 2, c = 4Find the values of a, b, and c. –(2) ± (2) 2 – 4(3)(4) 2(3) y = Substitute. –2 ± 4 – 48 6 –2 ± –44 6 = = –2 ± 2i 11 6 = Simplify. 5-8 Quick Check

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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? ALGEBRA 2 LESSON 5-8 The Quadratic Formula x 2 + (x + 1) 2 = 3 2 x 2 + x 2 + 2x + 1 = 9 2x 2 + 2x – 8 = 0Write in standard form. 17 – 1 2 The length of the shorter leg isunits. –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. 5-8

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(continued) ALGEBRA 2 LESSON 5-8 The Quadratic Formula – 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 2 5-8 Quick Check

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Determine the type and number of solutions of x 2 + 5x + 10 = 0. ALGEBRA 2 LESSON 5-8 The Quadratic Formula a = 1, b = 5, c = 10Find the values of a, b, and c. b 2 – 4ac = (5) 2 – 4(1)(10)Evaluate the discriminant. = 25 – 40Simplify. = –15 Since the discriminant is negative, x 2 + 5x + 10 = 0 has two imaginary solutions. 5-8 Quick Check

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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? ALGEBRA 2 LESSON 5-8 The Quadratic Formula Since the discriminant is negative, the equation 17 = –16t 2 + 25t + 6 has no real solution. The ball will hit the wall. h = –16t 2 + 25t + 6 17 = –16t 2 + 25t + 6Substitute 17 for h. 0 = –16t 2 + 25t – 11Write the equation in standard form. a = –16, b = 25, c = –11Find the values of a, b, and c. b 2 – 4ac = (25) 2 – 4(–16)(–11)Evaluate the discriminant. = 625 – 704 = –79 Simplify. Quick Check 5-8

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1.6x 2 + 19x + 8 = 02.x 2 – 2x – 11 = 0 3.x 2 – 2x – 15 = 04.3x 2 – 7x + 5 = 0 5.2x 2 – 5x + 7 = 06.–3x 2 – 14x – 8 = 0 7.4x 2 – 5x + 10 = 7x + 1 7 ± i 11 6 1 ± 2 3 ALGEBRA 2 LESSON 5-8 The Quadratic Formula –, – 8383 1212 –3, 5 –31; two, imaginary100; two, real 0; one, real Use the Quadratic Formula to solve each equation. Find the value of the discriminant for each quadratic equation. Tell how many different solutions each equation has and whether the solutions are real or imaginary. 5-8

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