Question 138197
{{{sqrt(y+3)=y-3}}}


Square both sides (since we are squaring the equation, we have to remember to check for extraneous roots at the end)
{{{y+3=(y-3)^2}}}
{{{y+3=y^2-6y+9}}}
{{{y^2-6y-y+9-3=0}}}
{{{y^2-7y+6=0}}}


The quadratic factors because {{{-1*-6=6}}} and {{{-1+(-6)=-7}}}


{{{(y-6)(y-1)=0}}}


Using the zero product rule:


{{{y-6=0}}} => {{{y=6}}}


or


{{{y-1=0}}} => {{{y=1}}}


Check for extraneous roots:
{{{sqrt(6+3)=6-3}}}
{{{sqrt(9)=3}}} True


{{{sqrt(1+3)=1-3}}}
{{{sqrt(4)<>-2}}} (radical sign means the positive square root by convention)
So {{{1}}} is an extraneous root.


The solution set is {6}