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

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Answer 1 · 2015-06-11T11:09:12

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

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.

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