# How do you solve x^2+2x-3<0?

Oct 23, 2016

${x}^{2} + 2 x - 3 < 0$ for $- 3 < x < 1$

#### Explanation:

${x}^{2} + 2 x - 3 < 0$
$\Rightarrow \left(x + 3\right) \left(x - 1\right) < 0$

$\textcolor{b l u e}{x + 3 < 0} \Rightarrow \textcolor{b l u e}{x < - 3}$
$\textcolor{b l u e}{x + 3 > 0} \Rightarrow \textcolor{b l u e}{x > - 3}$

$\textcolor{p u r p \le}{x - 1 < 0} \Rightarrow \textcolor{p u r p \le}{x < 1}$
$\textcolor{p u r p \le}{x - 1 > 0} \Rightarrow \textcolor{p u r p \le}{x > 1}$

Then,
$x + 3 < 0 \mathmr{and} x - 1 < 0 f \mathmr{and} - \infty < x < - 3$
then,$\left(x + 3\right) \left(x - 1\right) > 0$ $\textcolor{red}{R E J E C T E D}$

$x + 3 > 0 \mathmr{and} x - 1 < 0 f \mathmr{and} - 3 < x < 1$
then,$\left(x + 3\right) \left(x - 1\right) < 0$ $\textcolor{red}{A C C E P T E D}$

$x + 3 > 0 \mathmr{and} x - 1 > 0 f \mathmr{and} 1 < x < + \infty$
then,$\left(x + 3\right) \left(x - 1\right) > 0$ $\textcolor{red}{R E J E C T E D}$

Therefore,

${x}^{2} + 2 x - 3 < 0$ for $- 3 < x < 1$