Question 1007764

{{{(3k+5)^2-27=0}}}


{{{(3k+5)^2=27}}}........take a square root of both sides


{{{sqrt((3k+5)^2)=sqrt(27)}}}


{{{3k+5=sqrt(27)}}}


{{{3k=sqrt(27)-5}}}


{{{k=(sqrt(27)-5)/3}}}....since {{{27=3^2*3}}}, we have


{{{k=(sqrt(3^2*3)-5)/3}}}


{{{k=(3sqrt(3)-5)/3}}}


{{{k=cross(3)sqrt(3)/cross(3)-5/3}}}


{{{k=sqrt(3)-5/3}}}