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

Then $-2$ is less than $x+1$ is less than $2$ and hence $-1$ is less than $x$ is less than $1$.

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 open $(x+1 )^2$ subtract $4$ and then use wavy curve method 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.

Please help.

3 Answers 3

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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

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    @Sriram Cummaragunta: Why do you want to obtain $x$ between $-1$ and $+1$? – 2017-01-27
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    Never mind I got it.Thanks – 2017-01-30
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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,

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

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

$$x+1<2$$

$$AND$$

$$-x-1<-2$$

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    In the OR part the inequality evaluates to x>1 which does not satisfy the original condition.Perhaps that was an error? – 2017-01-21
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    @Sriram Cummaragunta See, that's what I'm saying. You almost always take the positive square root, hence the solution for the negative is mostly not given. – 2017-01-24
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    @MalayThe Dynamo: I think it should be *AND*, not *OR*. – 2017-01-27
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    @Bernard Yes. Thanks. – 2017-01-29
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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