How do you solve # sqrt(6x+1) = 3sqrt(x)-1#?

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Answer 1 · 2016-05-24T09:14:20

#x=0# or #x=4#

Squaring both sides of #sqrt(6x+1)=3sqrtx-1#, we get

#6x+1=9x+1-6sqrtx#

or #6sqrtx=9x+1-6x-1#

or #6sqrtx=3x#

or #2sqrtx=x# - now squaring this

#4x=x^2# or #x^2-4x=0#

or #x(x-4)=0#

Hence #x=0# or #x=4#