How do you solve ${(x - 3)}^{2} = 25$ $(x - 3)^2 = 25$ ?

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Answer 1 · 2016-04-13T18:10:57

$\Rightarrow x = 8 or x = - 2$ $=>x=8 or x=-2$

Take the square root of both sides giving

$\sqrt{{(x - 3)}^{2}} = \pm \sqrt{25}$ $sqrt( (x-3)^2) = +-sqrt(25)$

Technically it should be $\pm \sqrt{{(x - 3)}^{2}} = \pm \sqrt{25}$ $+-sqrt( (x-3)^2) = +-sqrt(25)$

But the same answer would be obtained by

$\sqrt{{(x - 3)}^{2}} = \pm \sqrt{25}$ $sqrt( (x-3)^2) = +-sqrt(25)$

$\Rightarrow x - 3 = \pm 5$ $=> x-3=+-5$

$\Rightarrow x = 3 \pm 5$ $=> x=3+-5$

$\Rightarrow x = 8 or x = - 2$ $=>x=8 or x=-2$

How do you solve (x - 3)^2 = 25(x−3)2=25(x - 3)^2 = 25?