Section 1.5 – Solve Quadratic Equations - McGraw-Hill Ryerson

) 8x 2 11x 5 5 0a 5 8, b 5 11, and c 5 5.________x 5 ​ b ____ ​ b 2 4ac ​2a _______________5 ​ (11) ______ ​ (11) 2 4(8)(5) ​_____ 2(8)5 ​___11 ​ 39 ​16Since the square root of a negative value is not a real number, there isno real solution to the quadratic equation.c) 4x 2 12x 9 5 0a 5 4, b 5 12, and c 5 9.________x 5 ​ b ____ ​ b 2 4ac ​2a________________5 ​ 12 _____ ​ 12 2 4(4)(9) ​​__2(4)5 ​___12 ​ 0 ​85 ​ _ 128 ​5 ​ 3_2 ​There is one solution because the value under the square root is zero.This means that there is exactly one root to the equation4x 2 12x 9 5 0.Example 3 shows that the value under the radical sign in the quadraticformula determines the number of solutions for a quadratic equation andthe number of zeros for the related quadratic function.Example 4Use the Discriminant to Determine the Number of SolutionsFor each quadratic equation, use the discriminant to determine thenumber of solutions.a) 2x 2 3x 8 5 0 b) 3x 2 5x 11 5 0 c) ​ 1_4 ​x2 3x 9 5 0Solutiona) 2x 2 3x 8 5 0a 5 2, b 5 3, and c 5 8.b 2 4ac 5 3 2 4(2)(8)5 9 645 73discriminant• the expressionb 2 — 4ac, the value ofwhich can be used todetermine the numberof solutions to aquadratic equationax 2 + bx + c = 0• When b 2 4ac > 0,there are two solutions.• When b 2 4ac 5 0,there is one solution.• When b 2 4ac < 0,there are no solutions.1.5 Solve Quadratic Equations • MHR 47