How do you solve #x- 6= \sqrt { 6x + 4}#?

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Answer 1 · 2017-05-11T10:24:46

#x- 6= \sqrt { 6x + 4}#

Taking squares in both sides

#(x-6)^2=6x+4#

#x^2-12x+36-6x-4=0#

#x^2-18x+32=0#

#x^2-16x-2x+32=0#

#(x-16)*(x-2)=0#

Because #x-6=sqrt(6x+4)>=0=>x>=6#

Hence the only acceptable solution is #x=16#