First, write the inequality in the form ax 2 + bx + c > 0.

1. - 2/3 3 Quadratic Inequality: Solving Algebraically EXAMPLE: Solve 3x2 7x > 6. First, write the inequality in the form ax2 + bx + c > 0. 3x2 7x 6 > 0 Next, solve the quadratic equation ax2 + bx + c = 0. (Here solve 3x2 7x 6 > 0.) The solutions are called critical values. The critical values are - 2/3 and 3. Next partition the real number line with the found critical values as shown.

2. x = - 10 x = 0 x = 10 ineq. is true ineq. is false ineq. is true - 2/3 3 Quadratic Inequality: Solving Algebraically For each interval replace the variable in the inequality with a number that belongs to that interval to determine if the inequality 3x2 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. Similarly, (- , -2/3) is a solution interval. Last, write the solution set as: (- , -2/3)  (3, ). Slide 2

3. Quadratic Inequality: Solving Algebraically Try to solve 2x2< 5 – 9x. The solution set is (- 5, 1/2 ). 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 3x2 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. Slide 3

4. Quadratic Inequality: Solving Algebraically END OF PRESENTATION Click to rerun the slideshow.