# How do you solve the inequality |6x^2-3|<7x?

May 1, 2016

$x \in \left(\frac{1}{3.} \frac{3}{2}\right)$

#### Explanation:

The given inequality is the combined inequality for the pair

$6 {x}^{2} - 3 < 7 x \mathmr{and} - \left(6 {x}^{2} - 3\right) < 7 x$.

So, $\left(2 x - 3\right) \left(3 x + 1\right) < 0 \mathmr{and} \left(2 x + 3\right) \left(3 x - 1\right) > 0$. These are equivalent to.

$\left(x - \frac{3}{2}\right) \left(x + \frac{1}{3}\right) < 0 \mathmr{and} \left(x + \frac{3}{2}\right) \left(x - \frac{1}{3}\right) > 0$.

Use that, if $a < b \mathmr{and} \left(x - a\right) \left(x - b\right) < 0$, then $x \in \left(a , b\right)$

and.if $a < b \mathmr{and} \left(x - a\right) \left(x - b\right) > 0$, then x < a and x > b#..

Accordingly, $x \in \left(- \frac{1}{3} , \frac{3}{2}\right)$ and x does not $\in \left[- \frac{3}{2} , \frac{1}{3}\right]$.
So, $x \in$ (intersection of these two intervals)$= \left(\frac{1}{3} , \frac{3}{2}\right)$.

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