How do you solve #3+ \sqrt { 2x - 3} = x#?

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Answer 1 · 2017-01-16T05:48:26

#3+sqrt(2x-3)=x => x in {2,6}#

#x=2 or x=6#

#3+sqrt(2x-3)=x#

#<=>#

#sqrt(2x-3)=x-3# Subtract 3 from both sides

#<=>#

square both sides, and use FOIL to find #(x-3)^2#

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

#=x^2-6x+9=2x-3#

#<=>#

#x^2-6x+12=2x# add 3 to both sides

#<=>#

#x^2-8x+12=0=x^2-(6+2)x+(-6)(-2)# factor

#<=>#

#(x-6)(x-2)=0 => x in{2,6}#

since the expression equals 0, x is either 2 or 6