Question

How do you solve the equation and identify any extraneous solutions for `sqrt(x+7) = x + 1`?

Answer 1

The term under the root must be `>=0` so `x>=-7`
The outcome must also be `>0` so `x>=-1`

Explanation:

Now we can square both sides:
`x+7=(x+1)^2=x^2+2x+1->`
`x^2+2x-x+1-7=x^2+x-6=0`
`(x+3)(x-2)=0->x=-3orx=2`

The only allowed solution is `x=2` as the other one does not meet the conditions at the start.