How do you solve #sqrt(x+7) = x + 1# and find any extraneous solutions?

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Answer 1 · 2018-04-19T23:00:37

I tried this:

Let us square both sides:

#(sqrt(x+7))^2=(x+1)^2#

#x+7=x^2+2x+1#

rearrange:

#x^2+x-6=0#

#x_(1,2)=(-1+-sqrt(1+24))/2=(-1+-5)/2#

giving:

#x_1=-3# that doesn't work;
#x_2=2#.