How do you solve #sqrt( x^2+7)=x+1#?
1 Answer
May 8, 2016
Explanation:
First square both sides, noting that squaring may introduce spurious solutions...
#x^2+7 = (x+1)^2 = x^2+2x+1#
Subtract
#7 = 2x+1#
Subtract
#6 = 2x#
Divide both sides by
#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#