How do you solve #sqrt[x+2]-sqrt[6x]=-1# and find any extraneous solutions?

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Answer 1 · 2018-02-20T01:16:20

#sqrt(x+2)=sqrt(6x)-1#

#(sqrt(x+2))^2=(sqrt(6x)-1)^2#

#x+2=6x-2sqrt(6x)+1#

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

#((-5x+1)/-2)^2=(sqrt(6x))^2#

#(25x^2-10x+1)/4=6x#

#(25x^2-10x+1)=24x#

#25x^2-34x+1=0#

#x=(34(+/-)sqrt(34^2-4*25*1))/(2*25)#

#x=0.03, 1.33# and 0.03 does not work when you plug it back into the equation to solve for x, so it is extraneous