How do you simplify #sqrt(n^2-8) = sqrt(2-3n)#?

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Answer 1 · 2016-04-16T21:47:27

#n={-5,2}#

#sqrt(n^2-8)=sqrt(2-3n)#

#(sqrt(n^2-8))^2=(sqrt(2-3n))^2#

#n^2-8=2-3n#

#n^2-8-2+3n=0#

#n^2+3n-10=0#

#(n+5)(n-2)=0#

#if (n+5)=0" then " n=-5#

#if(n-2)=0" then " n=2#

#n={-5,2}#