Question

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

Answer 1

`x < -4`

Explanation:

Note that `(x-2)^2 >= 0` for any real value of `x` and only takes the value `0` when `x=2`

When `x=2` we have:

`(x-2)^2/(x+4) = 0/6 = 0" "` 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 != 2`, so `(x-2)^2 > 0` and:

`(x-2)^2/(x+4) < 0`

So it is sufficient that `x < -4`

Answer 2

`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 `(x+4)`, the sign will not be flipped and we end up with:

`(x-2)^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 `(x+4)`, the sign will be flipped because we are multiplying by a negative number and we end up with:

`(x-2)^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 :-)