How do you solve $(x-5)^2=4$ ?

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Answer 1 · 2016-07-16T07:54:34

We must find which numbers $t$ satisfy $t^2=4$

If $t^2=4$ , then either $t=2$ or $t=-2$ .

But then we have from the given equation that either $(x-5)=2$ or $(x-5)=-2$ . Thus we have either:

$x=5+2$ , or $x=5-2$ . That is either:

$x=7$ , or $x=3$ , which are the solutions we are looking for.

How do you solve (x-5)^2=4(x-5)^2=4?