# If sqrt(x-1)=2, what is (x - 1)^2 ?

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Jim G. Share
Apr 22, 2018

${\left(x - 1\right)}^{2} = 16$

#### Explanation:

$\textcolor{b l u e}{\text{square both sides}}$

$\text{note that } \sqrt{a} \times \sqrt{a} = {\left(\sqrt{a}\right)}^{2} = a$

$\Rightarrow {\left(\sqrt{x - 1}\right)}^{2} = {2}^{2}$

$\Rightarrow x - 1 = 4$

$\textcolor{b l u e}{\text{square both sides}}$

${\left(x - 1\right)}^{2} = {4}^{2}$

$\Rightarrow {\left(x - 1\right)}^{2} = 16$

Then teach the underlying concepts
Don't copy without citing sources
preview
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

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VNVDVI Share
Apr 22, 2018

${\left(x - 1\right)}^{2} = 16$

#### Explanation:

Square both sides:

${\left(\sqrt{x - 1}\right)}^{2} = {\left(2\right)}^{2}$

This causes the radical to cancel out:

$\left(x - 1\right) = 4$

Square both sides, again:

${\left(x - 1\right)}^{2} = {4}^{2}$

${\left(x - 1\right)}^{2} = 16$

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