How do you solve $sqrt(x-7) + 5 = 11$ and find any extraneous solutions?

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Answer 1 · 2016-09-25T17:37:01

$x=43$

In the equation $sqrt(x-7)+5=11$ , we cannot have $x<7$ .

Now $sqrt(x-7)+5=11$

$hArrsqrt(x-7)=11-5=6$

or squaring both sides $x-7=6^2=36$

and $x=36+7=43$

(as it is not less than $7$ , it is not extraneous).

How do you solve sqrt(x-7) + 5 = 11sqrt(x-7) + 5 = 11 and find any extraneous solutions?