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Copyright 2013, 2009, 2005, 2002 Pearson, Education, Inc.

Solving Quadratic Equations by the Quadratic Formula 8.2 Solving Quadratic Equations by the Quadratic Formula

The Quadratic Formula Another technique for solving quadratic equations is to use the quadratic formula. The formula is derived from completing the square of a general quadratic equation.

The Quadratic Formula Quadratic Formula A quadratic equation written in the form ax2 + bx + c = 0 has the solutions

Example Solve 3n2 + n – 3 = 0 by the quadratic formula. a = 3, b = 1, c = –3 5

Example Solve 11n2 – 9n = 1 by the quadratic formula. a = 11, b = – 9, c = – 1 6

Example Solve x2 + x – = 0 by the quadratic formula. x2 + 8x – 20 = 0 Multiply both sides by 8. a = 1, b = 8, c = –20 7

Example Solve x(x + 6) = –30 by the quadratic formula. a = 1, b = 6, c = 30 There is no real solution. 8

The Discriminant The expression under the radical sign in the formula (b2 – 4ac) is called the discriminant. b2 – 4ac Number and Type of Solutions Positive Two real solutions Zero One real solution Negative Two complex but not real solutions

Example Use the discriminant to determine the number and type of solutions for the following equation. 5 – 4x + 12x2 = 0 a = 12, b = –4, and c = 5 b2 – 4ac = (–4)2 – 4(12)(5) = 16 – 240 = –224 There are no real solutions.

Example Use the discriminant to determine the number and type of solutions for the following equation. 2x2 – 7x – 4 = 0 a = 2, b = –7, and c = –4 b2 – 4ac = (–7)2 – 4(2)(–4) = 81 There are two real solutions.

Example At a local university, students often leave the sidewalk and cut across the lawn to save walking distance. Given the diagram below of a favorite place to cut across the lawn, approximate to the nearest foot how many feet of walking distance a student saves by cutting across the lawn instead of walking on the sidewalk. Continued 13

Example (cont) Use the Pythagorean theorem. Continued

Example (cont) Use the Pythagorean theorem. Continued

Example (continued) Use the Pythagorean theorem. x + (x + 20) ≈ 24 + (24 + 20) = 68 feet A person saves about 68 – 50 or 18 feet of walking distance by cutting across the lawn.