Question

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

Answer 1

Two cases

`a. -(6x^2 - 3) < 7x -> f(x) = -6x^2 - 7x + 3 < 0` (1)

`b. (6x^2 - 3) < 7x -> f(x) = 6x^2 - 7x - 3 < 0` (2)
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: `p = p'/a = -2/-6 = 1/3`. and `q' = q/a = 9/-6 = -3/2`

===============|-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
`p' = -2 -> p = -2/6 = -1/3`
`q' = 9 -> q = 9/6 = 3/2.`

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