# How do you solve (x-2)^2/(x+4) < 0 ?

Sep 14, 2017

#### Answer:

$x < - 4$

#### Explanation:

Note that ${\left(x - 2\right)}^{2} \ge 0$ for any real value of $x$ and only takes the value $0$ when $x = 2$

When $x = 2$ we have:

${\left(x - 2\right)}^{2} / \left(x + 4\right) = \frac{0}{6} = 0 \text{ }$ which does not satisfy the inequality.

The only circumstance under which the rational expression is negative is if the denominator is negative:

$x + 4 < 0$

Subtracting $4$ from both sides, that means we need:

$x < - 4$

Note that when $x < - 4$, we have $x \ne 2$, so ${\left(x - 2\right)}^{2} > 0$ and:

${\left(x - 2\right)}^{2} / \left(x + 4\right) < 0$

So it is sufficient that $x < - 4$

Sep 14, 2017

#### Answer:

$x < - 4$

#### Explanation:

George's explanation is very cool and on the money! Here's a different way to look at it.

When dealing with inequalities, if you multiply or divide by a negative number we have to flip the sign. In this case, we don't know if the denominator, $x + 4$, is negative or positive so we have two cases.

If we assume that $x + 4 > 0$ or in other words it's positive, $x > - 4$.

Going back to the inequality, if we now multiply both sides by $\left(x + 4\right)$, the sign will not be flipped and we end up with:

${\left(x - 2\right)}^{2} < 0$

If you square a quantity, it is always positive, so the inequality above is invalid.

If you look at the second case and assume $x + 4 < 0$ or in other words it's negative, $x < - 4$.

Going back to the inequality, if we now multiply both sides by $\left(x + 4\right)$, the sign will be flipped because we are multiplying by a negative number and we end up with:

${\left(x - 2\right)}^{2} > 0$

Since we just, found $x < - 4$, we know that the inequality above will be satisfied. It will only NOT be satisfied if $x = 2$ because then it will be equal to zero.

Here's a good video:

https://www.khanacademy.org/math/algebra-home/alg-rational-expr-eq-func/alg-rational-inequalities/v/rational-inequalities

It's a little bit tricky :-)