Table of Contents First, write the inequality in the formax 2 + bx + c > 0. Quadratic Inequality: Solving Algebraically EXAMPLE: Solve 3x 2  7x > 6. 3x 2  7x  6 > 0 The critical values are - 2/3 and 3. Next partition the real number line with the found critical values as shown. Next, solve the quadratic equation ax 2 + bx + c = 0. (Here solve 3x 2  7x  6 > 0.) The solutions are called critical values. - 2/3 3

Table of Contents Quadratic Inequality: Solving Algebraically Slide 2 For each interval replace the variable in the inequality with a number that belongs to that interval to determine if the inequality 3x 2  7x > 6 is true or false. Theory says that since x = 10 made the inequality true, so will any other number in that interval so (3,  ) is part of the solution set. Last, write the solution set as: (- , -2/3)  (3,  ). Similarly, (- , -2/3) is a solution interval. x = - 10 x = 0x = 10 ineq. is true ineq. is false ineq. is true - 2/3 3

Table of Contents The solution set is (- 5, 1/2 ). Quadratic Inequality: Solving Algebraically Try to solve 2x 2 < 5 – 9x. Slide 3 Note: Sometimes you will find only one critical value. In this case the number line is partitioned into 2 intervals so there are only two intervals to test. Sometimes you will find no critical values (they must be real) and in this case the number line is not partitioned, however think of it as consisting of one interval so there is still one interval to test. Also note that if solving 3x 2  7x  6 (instead of > as in the example on the preceding slides), the solution set is (- , -2/3]  [3,  ). In this case, since the critical values also make the inequality true, they are included in the solution set.

Table of Contents Quadratic Inequality: Solving Algebraically