Taking square roots in inequalities

Question

I ran into a small problem while taking the square root on both sides of an inequality. Suppose it is given that $(x+1)^2 < 4$.

Then, after taking the square root of both sides, $-2 < x+1 < 2$ and hence $-1 < x < 1$.

But on further analysis it appears that the solution of $x$ where $x$ ranges from $-3$ to $1$ is not obtained.

I know that I can expand $(x+1)^2$, subtract $4$, and then use wavy curve method (i.e., the method of intervals) to get the complete solution.

However my confusion lies in why the solutions have not been obtained by taking the square root on both sides of the inequality.

Answer 1

Interpret this inequation in terms of distance: $$(x+1)^2<4\iff\sqrt{(x+1)^2}=\lvert x+1\rvert<2.$$ Now $\lvert x+1\rvert$ is the distance betwen $x$ and $-1$, so $$\lvert x+1\rvert<2\iff -1-2<x<-1+2.$$

Answer 2

Your problem is that taking square roots gives you two alternative answers with same absolutes but different signs. Generally, if taking square root, we take only the positive root. Here,

$$(x+1)^2<4$$

$$\sqrt{(x+1)^2}<\sqrt4$$

$$x+1<2$$

$$AND$$

$$-x-1<-2$$

Answer 3

You're doing nothing wrong involving square roots; your problem is with subtraction. You are correct that $(x+1)^2<4$ is equivalent to $$-2<x+1<2.$$ However, it is not correct to then conclude that $-1<x<1$. Rather, you must subtract $1$ from each part of the inequality, giving $$-2-1<(x+1)-1<2-1$$ or $$-3<x<1.$$