How do you solve #sqrt( x^2+7)=x+1#?

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Answer 1 · 2016-05-08T12:22:59

#x=3#

First square both sides, noting that squaring may introduce spurious solutions...

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

Subtract #x^2# from both ends to get:

#7 = 2x+1#

Subtract #1# from both sides to get:

#6 = 2x#

Divide both sides by #2# and transpose to get:

#x = 3#

Check that this is a solution of the original equation:

#sqrt(x^2+7) = sqrt(3^2+7) = sqrt(9+7) = sqrt(16) = 4 = 3+1 = x+1#