How do you solve #x - 9 = sqrt(x - 3)#?

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Answer 1 · 2016-07-01T10:15:29

#x=12#

First you fix the domain of the solutions:

#x-3>=0 and x-9>=0#

that's #x>=9#

then you square both members:

#(x-9)^2=(sqrt(x-3))^2#

that's

#x^2-18x+81=x-3#

#x^2-18x+81-x+3=0#

#x^2-19x+84=0#

#x=(19+-sqrt(19^2-4*84))/2#

#x=(19+-5)/2#

#x=7 and x=12#

but only x=12 belongs to the calculated domain