Quadratic Inequalities

In this case, these are the solutions to $\sqrt{x^2}$. Remember that there is a positive and a negative square root, so the critical values are $\sqrt{9}=3$or$\sqrt{9}=-3$.

If $x^2<9$then $x$ must be less than $3$. This is because $3^2=9$, so anything less than 9 must have a smaller square root.

If $x^2<9$, then $x$must be greater than $-3$. $-3^2=9$. Therefore, anything less than 9 should have a square root closer to 0 (and therefore greater).

$-3 < x < 3$

To do this, we need to replace the $>$ sign with an $=$ sign:$x^2 + 3x + 2=0$

$x=-1$ and $x=-2$. We can use the critical values to identify the solutions on a graph.