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

2 Answers
Sep 14, 2017

Answer:

#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#

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 #(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 :-)