# How do you solve 4t ^ { 2} > 12?

Mar 10, 2018

$t > \sqrt{3}$ OR $t < - \sqrt{3}$

#### Explanation:

We have the inequality:

$4 {t}^{2} > 12$

1) Divide both sides by $4$:

$\frac{4 {t}^{2}}{4} > \frac{12}{4}$

$\implies {t}^{2} > 3$

2) Take the square root of both sides:

$\sqrt{{t}^{2}} > \sqrt{3}$

$\implies t > \sqrt{3}$

BUT square roots have two solutions - positive or negative. You know that, when we divide by a negative, we change the sign of the inequality. So:

$t > \sqrt{3}$ OR $t < - \sqrt{3}$

Mar 10, 2018

The solution is $t \in \left(- \infty , - \sqrt{3}\right) \cup \left(\sqrt{3} , + \infty\right)$

#### Explanation:

Let's rearrange the inequality

$4 {t}^{2} > 12$

$4 {t}^{2} - 12 > 0$

Factorise,

$4 \left({t}^{2} - 3\right) > 0$

$4 \left(t + \sqrt{3}\right) \left(t - \sqrt{3}\right) > 0$

Let $f \left(t\right) = 4 \left(t + \sqrt{3}\right) \left(t - \sqrt{3}\right)$

Now build a sign chart

$\textcolor{w h i t e}{a a a a}$$t$$\textcolor{w h i t e}{a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a}$$- \sqrt{3}$$\textcolor{w h i t e}{a a a a}$$\sqrt{3}$$\textcolor{w h i t e}{a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$t + \sqrt{3}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$t - \sqrt{3}$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(t\right)$$\textcolor{w h i t e}{a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(t\right) > 0$ when $t \in \left(- \infty , - \sqrt{3}\right) \cup \left(\sqrt{3} , + \infty\right)$

graph{4x^2-12 [-25.65, 25.66, -12.83, 12.84]}