# What is the solution set for -x^2 + 2x > -3?

Aug 31, 2015

$x \in \left(- 1 , 3\right)$

#### Explanation:

Start by getting all the terms on one side of the inequality. You can do that by adding $3$ to both sides

$- {x}^{2} + 2 x + 3 > - \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}$

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

Next, make the quadratic equal to zero in order to find its roots. This will help you factor it. Use the quadratic formula to calculate ${x}_{1 , 2}$.

$- {x}^{2} + 2 x + 3 = 0$

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{{2}^{2} - 4 \cdot \left(- 1\right) \cdot \left(3\right)}}{2 \cdot \left(- 1\right)}$

${x}_{1 , 2} = \frac{- 2 \pm \sqrt{16}}{\left(- 2\right)}$

${x}_{1 , 2} = \frac{- 2 \pm 4}{\left(- 2\right)} = \left\{\begin{matrix}{x}_{1} = \frac{- 2 - 4}{\left(- 2\right)} = 3 \\ {x}_{2} = \frac{- 2 + 4}{\left(- 2\right)} = - 1\end{matrix}\right.$

This means that you can rewrite the quadratic as

$- \left(x - 3\right) \left(x + 1\right) = 0$

Your inequality will be equivalent to

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

In order for this inequality to be true, you need one of the two terms to be positive and the other negative, or vice versa.

Your first two conditions will be

$x - 3 > 0 \implies x > 3$

and

$x + 1 < 0 \implies x < - 1$

Since you can't have values of $x$ that are both greater than $3$ and smaller than $\left(- 1\right)$, this possibility is eliminated.

The other conditions wll be

$x - 3 < 0 \implies x < 3$

and

$x + 1 > 0 \implies x > - 1$

This time, these two intervals will produce a valid solution set. For any value of $x$ that is greater than $\left(- 1\right)$ and smaller than $3$, this product

$\left(x - 3\right) \cdot \left(x + 1\right) < 0$

which means that

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

The solution set for this inequality will thus be $x \in \left(- 1 , 3\right)$.