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

May 10, 2015

Two cases

$a . - \left(6 {x}^{2} - 3\right) < 7 x \to f \left(x\right) = - 6 {x}^{2} - 7 x + 3 < 0$ (1)

$b . \left(6 {x}^{2} - 3\right) < 7 x \to f \left(x\right) = 6 {x}^{2} - 7 x - 3 < 0$ (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: $p = p \frac{'}{a} = - \frac{2}{-} 6 = \frac{1}{3}$. and $q ' = \frac{q}{a} = \frac{9}{-} 6 = - \frac{3}{2}$

===============|-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
$p ' = - 2 \to p = - \frac{2}{6} = - \frac{1}{3}$
$q ' = 9 \to q = \frac{9}{6} = \frac{3}{2.}$

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