How do you solve #sqrt(5x-9)-1=sqrt( 3x-6)# and find any extraneous solutions?

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Answer 1 · 2018-07-06T11:32:43

#x=5# or #x=2#

We have #x\geq 2# since the radicand must be non negative.
Writing your equation in the form
#sqrt(5x-9)=sqrt(3x-6)+1#

squaring we get

#5x-9=3x-6+1+2sqrt(3x-6)#
collecting like Terms and isolating the square root

#x-2=sqrt(3x-6)#
since #x\geq 2# we can square again

#x^2-7x+10=0#
solving this quadratic equation by the quadratic formula we get

#x_(1,2)=7/2pmsqrt(49/4-40/4)#

so

#x_1=5#

#x_2=2#
checking both Solutions we get that both fulfill the equation above.

Answer 2 · 2018-07-06T19:05:30

#x_1=2# and #x_2=5#

#sqrt(5x-9)-1=sqrt(3x-6)#

#5x-9+1-2sqrt(5x-9)=3x-6#

#2x-2=2sqrt(5x-9)#

#x-1=sqrt(5x-9)#

#(x-1)^2=5x-9#

#x^2-2x+1=5x-9#

#x^2-7x+10=0#

#(x-2)*(x-5)=0#

So, #x_1=2# and #x_2=5# are solutions of this equation.