# How do you solve #abs(6x^2-3)<7x#?

##### 1 Answer

Two cases

a. Solve quadratic equation (1) by the Transforming Method (Google, Yahoo Search)

Transformed equation: x^2 - 7x - 18 = 0 (a.c = -18)

Compose factor pairs of -18. Proceed: (-1, 18)(-2, 9). This last sum is 9 - 2 = 7 = -b. Then the 2 real roots p' and q' are: -2 and 9. The 2 real roots of original equation (1) are:

===============|-3/2 -------------0 -------|1/3 ==============

On the number line, use test point x= 0. f(0) = 3 < 0.. Not true. Then the solution set are the 2 rays.

b. Solve equation (2). f(x) = 6x^2 - 7x - 3 < 0 (2) Use same method to get

--------------------------|-1/3============|3/2---------------

The solution set is the open interval (-1/3, 3/2).