If $\sqrt{x - 1} = 2$ $sqrt(x-1)=2$ , what is ${(x - 1)}^{2}$ $(x - 1)^2$ ?

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If $\sqrt{x - 1} = 2$ $sqrt(x-1)=2$ , what is ${(x - 1)}^{2}$ $(x - 1)^2$ ?

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Answer 1 · 2018-04-22T18:24:04

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

Explanation:

Square both sides:

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

This causes the radical to cancel out:

$(x - 1) = 4$ $(x-1)=4$

Square both sides, again:

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

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

Answer 2 · 2018-04-22T18:30:02

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

Explanation:

$square both sides$ $color(blue)"square both sides"$

$note that \sqrt{a} \times \sqrt{a} = {(\sqrt{a})}^{2} = a$ $"note that "sqrtaxxsqrta=(sqrta)^2=a$

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

$\Rightarrow x - 1 = 4$ $rArrx-1=4$

$square both sides$ $color(blue)"square both sides"$

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

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

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If $\sqrt{x - 1} = 2$ $sqrt(x-1)=2$ , what is ${(x - 1)}^{2}$ $(x - 1)^2$ ?

2 Answers

Explanation:

Explanation:

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If $\sqrt{x - 1} = 2$ $sqrt(x-1)=2$ , what is ${(x - 1)}^{2}$ $(x - 1)^2$ ?