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

$x \ge 1 , x \le - 3$

#### Explanation:

The key to questions like this one is to solve it in the same way that you solve an equality (so one with an $=$ sign), with two caveats:

• that if you multiply or divide by a negative number, you flip the greater than sign, and
• we'll be setting zones to see what is valid and what isn't.

So starting with the original:

${x}^{2} + 2 x - 3 \ge 0$

we can factor this into:

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

and solve for x:

$x \ge - 3$ and $x \ge 1$

Ok, so we know that we have 3 zones: $x \ge 1$, $x \le - 3$, and the region between 1 and $- 3$. Let's test $x = - 2$ to see if the middle region is valid:

${x}^{2} + 2 x - 3 \ge 0$

${\left(- 2\right)}^{2} + 2 \left(- 2\right) - 3 \ge 0$

$4 - 4 - 3 \ge 0$

$- 3 \ge 0$ - nope.

And let's test $x = - 4$ to test the left-most region:

${x}^{2} + 2 x - 3 \ge 0$

${\left(- 4\right)}^{2} + 2 \left(- 4\right) - 3 \ge 0$

$16 - 8 - 3 \ge 0$

$5 \ge 0$ - yes.

And we can test the right-most region with $x = 2$:

${x}^{2} + 2 x - 3 \ge 0$

${2}^{2} + 2 \left(2\right) - 3 \ge 0$

$4 + 4 - 3 \ge 0$

$5 \ge 0$ - yes.

So the answer in notation is $x \ge 1 , x \le - 3$. The graph is below (on a number line, this will be solid lines on the x-axis in the shaded area starting from a solid point at x=1 and going right and another from x=-3 and going left.)

graph{x^2+2x-3>=0 [-10, 10, -1, 1]}