# How do you solve and write the following in interval notation: 6x^2+3x<4-2x?

May 19, 2016

Any x, sans $- \frac{4}{3} \mathmr{and} \frac{1}{2}$.

#### Explanation:

A rearrangement gives

$6 \left({x}^{2} + \left(\frac{5}{6}\right) x - \frac{2}{3}\right) < 0$.

So, ${x}^{2} + \left(\frac{5}{6}\right) x - \frac{2}{3}$

$= {\left(x + \frac{5}{12}\right)}^{2} - {\left(\frac{11}{12}\right)}^{2}$

$= \left(\left(x + \frac{5}{12}\right) - \frac{11}{12}\right) \left(\left(x + \frac{5}{12}\right) + \frac{11}{12}\right)$

$= \left(x - \frac{1}{2}\right) \left(x + \frac{4}{3}\right) < 0$

For the product to be negative, the factors should be of opposite signs.

So, $x > \frac{1}{2} \mathmr{and} < - \frac{4}{3.}$ Also, $x < \frac{1}{2} \mathmr{and} > - \frac{4}{3}$.

Thus, the answer is any x, sans $\frac{1}{2} \mathmr{and} - \frac{4}{3}$..