# How do you solve 4x^2-9>0 using a sign chart?

Jul 6, 2017

Solution : $x < - \frac{3}{2} \mathmr{and} x > \frac{3}{2}$ , In interval notation $\left(- \infty , - \frac{3}{2}\right) \cup \left(\frac{3}{2} , \infty\right)$

#### Explanation:

$4 {x}^{2} - 9 > 0 \mathmr{and} {\left(2 x\right)}^{2} - {3}^{2} > 0 \mathmr{and}$(2x+3)(2x-3) >0 

Critical points are $2 x + 3 = 0 \mathmr{and} x = - \frac{3}{2} \mathmr{and} 2 x - 3 = 0 \mathmr{and} x = \frac{3}{2}$

Sign chart:

When x < -3/2 ; (2x+3)(2x-3)  sign is $\left(-\right) \cdot \left(-\right) = + \therefore > 0$

When  -3/2 < x < 3/2 ; (2x+3)(2x-3)  sign is $\left(+\right) \cdot \left(-\right) = - \therefore < 0$

When x > 3/2 ; (2x+3)(2x-3) # sign is $\left(+\right) \cdot \left(+\right) = + \therefore > 0$

So Solution is $x < - \frac{3}{2} \mathmr{and} x > \frac{3}{2}$ , In interval notation $\left(- \infty , - \frac{3}{2}\right) \cup \left(\frac{3}{2} , \infty\right)$ [Ans]