Question

If $\sqrt{3}y=\sqrt{72+y^2}$, then find the value of $y.$

Answer 1

$\sqrt{3}y=\sqrt{72+y^2}$

For $y=?$

squaring both sides

$(\sqrt{3}y{)}^2=72+y^2$

$3y^2=72+y^2$

$2y^2=72$

$y^2=36$

$y=\pm 6$