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

Feb 21, 2016

First calculate the zeros of the polynom:

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

$x = \frac{- 2 \pm \sqrt{16}}{2}$

$x = \frac{- 2 \pm 4}{2}$

$x = - 3 \mathmr{and} x = 1$

So the inequality can be simplified into:

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

This will happen when one of the factors is positive and the other negative, so the solution will be:

$x \in \left[- 3 , 1\right]$

Feb 22, 2016

Closed Interval [-3, 1]

#### Explanation:

$f \left(x\right) = {x}^{2} + 2 x - 3 \le 0$
First , solve f(x) = 0.
Since a + b + c = 0, use shortcut. One real root is 1 and the other is c/a = -3.
Use the algebraic method to solve f(x) <= 0. Between the 2 real roots (x-intercepts), f(x) < 0. (The parabola graph is below the x-axis because a > 0)).
Solution set: closed interval [-3, 1].
The 2 end points ( -3 and 1) are included in the solution set.