# How do I use a sign chart to solve 2x^2-7x> -3?

Apr 4, 2018

Solution: $x < 0.5 \mathmr{and} x > 3 \mathmr{and} \left(- \infty , 0.5\right) \cup \left(3 , \infty\right)$

#### Explanation:

$2 {x}^{2} - 7 x > - 3 \mathmr{and} 2 {x}^{2} - 7 x + 3 > 0$ or

$2 {x}^{2} - 6 x - x + 3 > 0$ or

$2 x \left(x - 3\right) - 1 \left(x - 3\right) > 0$ or

$\left(x - 3\right) \left(2 x - 1\right) > 0$

Critical points:

$\left(x - 3\right) \left(2 x - 1\right) = 0$ . Critical points are $2 x - 1 = 0 \therefore x = \frac{1}{2}$

and $x - 3 = 0 \therefore x = 3 \therefore$ critical points are $\frac{1}{2} , 3$

Sign chart:

When $x < \frac{1}{2}$ sign of $\left(x - 3\right) \left(2 x - 1\right)$ is  (-) * (-) = (+) ; > 0

When $\frac{1}{2} < x < 3$ sign of $\left(x - 3\right) \left(2 x - 1\right)$ is  (-) * (+) = (-) ; < 0

When $x > 3$ sign of $\left(x - 3\right) \left(2 x - 1\right)$ is  (+) * (+) = (+) ; > 0

Solution: $x < 0.5 \mathmr{and} x > 3 \mathmr{and} \left(- \infty , 0.5\right) \cup \left(3 , \infty\right)$ [Ans]