Question

Find the positive root of $\sqrt{3x^2+6}=9$

Answer 1

Consider, $\sqrt{3x^2+6}=9$

squaring on both sides

$3x^2+6=81$

$3x^2=81-6$

$3x^2=75$

$x^2=\frac{75}{3}$

$x^2=25$

$x=\sqrt{25}$

$x=\pm 5$

But positive root:

$x=5$