How do you solve ${(x - 5)}^{2} = 9$ $(x-5)^2=9$ ?

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Answer 1 · 2016-07-25T10:47:54

Only two numbers are such that raised to the square $= 9$ $=9$

If a number to the square $= 9$ $=9$ , it is either $3$ $3$ or $- 3$ $-3$ . So, either:

$x - 5 = 3$ $x-5=3$ , or $x - 5 = - 3$ $x-5=-3$ . Then, it is either $x = 3 + 5$ $x=3+5$ or $x = - 3 + 5$ $x=-3+5$

The two solutions are then $x = 8$ $x=8$ and $x = 2$ $x=2$

How do you solve (x−5)2=9(x-5)^2=9?