Question

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

Answer 1

`x=43`

Explanation:

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).