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

1 Answer
MeneerNask
Jun 11, 2015

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.

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