Quadratic inequality (Sign Reversal?)

Question

I have the following inequality

$\ (2x-3)^2-9>7$

I can reduce it down to

$\ 2x-3>±4$

Now here is where I encounter a problem. Apparently the next step is

$\ 2x-3>4 ~OR~ 2x-3<-4 $

Why is the inequality sign is reversed in the second inequality?

Answer 1

You have $$(2x-3)^2>16$$ and you attempt to take square root both sides. The second inequality is wrong.

The sign is preserved when you apply a function that is monotonically non-decreasing to both sides (did you check this condition?)

When we apply the square root function to both sides of inequalities, we have

$$\left| 2x-3 \right|>4$$

where the absolute value comes from we are not sure if $2x-3$ is nonnegative.

You just have to solve this inequality to obtain your solution.

Answer 2

The intermediate step is $$ (2x-3)^2>16.\tag1 $$ Inequality (1) is saying that something (namely $2x-3$), when squared, exceeds 16. The something can be either positive or negative. If positive, it must exceed 4. If negative, it has to be less than $-4$. Now continue with these two cases.