How do you solve #x/(3 sqrt (2x-3)) - 1/ (sqrt (2x-3))=1/3# and find any extraneous solutions?

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Answer 1 · 2018-04-09T09:53:45

#x=6#

#x/(3sqrt(2x-3))-1/sqrt(2x-3)=1/3#

#(x-3)/(3sqrt(2x-3))=1/3#

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

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

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

#x^2-8x+12=0#

#(x-2)*(x-6)=0#

So #x_1=2#, #x_2=6#. But #x=2# is extraneous solution. Hence #x=6# is solution of this equation.