How do you solve #sqrt(3x-3)-sqrt(x+12)=0#?

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Answer 1 · 2016-07-06T06:23:33

The soln. is #x=15/2#

We notice that the eqn. will be meaningful (in #RR#) iff #(3x-3)>=0, &, x+12>=0,# i.e., #x>=1 & x>=-12#, equivalently, #x>=1...........(1)#

Given that, #sqrt(3x-3)-sqrt(x+12)=0.#

#rArr sqrt(3x-3)=sqrt(x+12)#

#rArr (sqrt(3x-3))^2=(sqrt(x+12))^2#

#rArr3x-3=x+12#

#rArr 2x=15#

#rArr x=15/2#

This soln. satisfy the given eqn. and #(1)# as well.

Hence, the soln. is #x=15/2#

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