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

Sep 20, 2016

Either $y < - 2$ or $y > 2$

#### Explanation:

As ${y}^{2} - 4 > 0$, factorizing we have $\left(y + 2\right) \left(y - 2\right) > 0$

From this we know that the product $\left(y + 2\right) \left(y - 2\right)$ is positive. It is apparent that sign of binomials $\left(y + 2\right)$ and $y - 2$ will change around the values $- 2$ and $2$ respectively. In sign chart we divide the real number line using these values, i.e. below $- 2$, between $- 2$ and $2$ and above $2$ and see how the sign of ${y}^{2} - 4$ changes.

Sign Chart

$\textcolor{w h i t e}{X X X X X X X X X X X} - 2 \textcolor{w h i t e}{X X X X X} 2$

$\left(y + 2\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left(y - 2\right) \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

$\left({y}^{2} - 4\right) \textcolor{w h i t e}{X X X} + i v e \textcolor{w h i t e}{X X X} - i v e \textcolor{w h i t e}{X X X X} + i v e$

It is observed that ${y}^{2} - 4 > 0$ when either $y < - 2$ or $y > 2$, which is the solution for the inequality.