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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?

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After seeing it happen many times, I can't help reminding you to be sure you're not confusing $\sqrt{3x}$ with $\sqrt{3}x$, which are two quite different things. –  Michael Hardy Apr 12 '13 at 15:42
    
Do you see that $\dfrac {3 \sqrt 3}{\sqrt 3} = 3 \sqrt {\dfrac{3}{3}} = 3 \sqrt 1 = 3$? –  amWhy Apr 12 '13 at 15:47
    
I would immediately square both sides and get rid of the square root. –  Stefan Smith Apr 12 '13 at 23:56
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2 Answers

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: $$(3)^2 = \left(\sqrt x\right)^2 \quad \iff \quad 9 = x$$

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You have $$ \sqrt{x} = \frac{3\sqrt{3}}{\sqrt{3}}. $$ Note here that $x$ must be positive. You can now simply square both sides: $$ x = (\sqrt{x})^2 = \left(\frac{3\sqrt{3}}{\sqrt{3}}\right)^2. $$ Left for you to do is to simplify things.

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