How do you solve #x-3 = sqrt(x+27)#?

1 Answer
Feb 25, 2016

Square both sides to get a quadratic to solve, then find which solution of the quadratic is a solution of the original equation.

Explanation:

Start by squaring both sides, but note that this typically introduces spurious solutions. That is, the resulting equation may have solutions that the original does not.

#x+27 = (x-3)^2=x^2-6x+9#

Subtract #x+27# from both ends to get:

#0 = x^2-7x-18 = (x-9)(x+2)#

(I found this factorisation by finding a pair of factors #9#, #2# of #18#, which differ by #7#)

This has solutions #x=9# and #x=-2#

The value #x=-2# is not a solution of the original equation, since if #x=-2# then:

#x-3 = -2-3 = -5 != 5 = sqrt(-2+27) = sqrt(x+27)#

The value #x=9# is a solution of the original equation, since:

#x-3 = 9-3 = 6 = sqrt(36) = sqrt(9+27) = sqrt(x+27)#