If I have one square root that involves a coefficient, and one square root that doesn't, how do I solve for $x$?
$$3\sqrt3 = \sqrt{3x} $$
I tried to simplify it like:
$$ \frac{3\sqrt3} {\sqrt3} = \sqrt x$$
But ran into a dead end...what do I do next?
Note that: $$\dfrac {3 \sqrt 3}{\sqrt 3} = 3 \sqrt {\dfrac{3}{3}} = 3 \sqrt 1 = 3$$
But we can think of approaching the original equation by first noting: $$3\sqrt3 = \sqrt{3x} \iff 3\sqrt 3 = \sqrt 3 \times \sqrt x$$
and "canceling" $\;\sqrt 3\;$ from each side of the equation.
$$3\sqrt 3 = \sqrt 3 \sqrt x \quad \iff \quad 3 = \sqrt x$$
Now simply square both sides of the equation: $$(3)^2 = \left(\sqrt x\right)^2 \quad \iff \quad 9 = x$$