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How would I simplify the following two radicals.


For my answer I got $\frac{X^2}{50X}$ but I am not sure if this is correct.

My second question is how would I simplify $\sqrt{\frac{1}{12}}$

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It sounds like you could benefit from reviewing the laws of radicals again. – Asaf Karagila Jun 13 '12 at 20:43
I added LaTeX to your equations. Please correct them if this is not what you meant, I didn't completely understand. – Argon Jun 13 '12 at 20:47

No, your simplification is incorrect.

The first radical only makes sense if $x\geq 0$. Assuming this is the case, remember that:

  1. If $a$ and $b$ are both nonnegative, then $\sqrt{ab}=\sqrt{a}\sqrt{b}$, and $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$.

  2. $(\sqrt{a})^2 = a$ for any $a\geq 0$.


$$\sqrt{\frac{x^3}{50}} = \frac{\sqrt{x^3}}{\sqrt{50}} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}} = \frac{x\sqrt{x}\sqrt{2}}{5\sqrt{2}\sqrt{2}} = \frac{x\sqrt{2x}}{10}.$$

There are other ways of expressing it as well.

For the second, $12 = 4\times 3$, so $$\sqrt{12}=\sqrt{4}\sqrt{3} = 2\sqrt{3}.$$ Can you take it from there?

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For your first one,

$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^2}\sqrt{x}}{\sqrt{25}\cdot\sqrt{2}} = \frac{x\sqrt{x}}{5\sqrt{2}}$$

At this point, you could multiply top and bottom through by $\sqrt{2}$ (to make the $\sqrt{2}$ part of the bottom "go away"):

$$\frac{x\sqrt{x}\cdot\sqrt{2}}{5\sqrt{2}\cdot\sqrt{2}} = \frac{x\sqrt{2x}}{10}$$

For the second one, $$\sqrt\frac{1}{12} = \frac{1}{\sqrt{12}} = \frac{1}{\sqrt{4}\cdot\sqrt{3}} = \frac{1}{2\sqrt{3}}$$

From here, you could opt to multiply top and bottom through by $2\sqrt{3}$ as shown below (which is called rationalizing the denominator - to get rid of square root terms in the denominator.)

$$\frac{1\cdot{2\sqrt{3}}}{2\sqrt{3}\cdot{2\sqrt{3}}} = \frac{2\sqrt{3}}{4\cdot\sqrt{9}} = \frac{\sqrt{3}}{2\cdot{3}} = \frac{\sqrt{3}}{6}$$

When dealing with problems like these, it is helpful to split what is under the square root into two separate square root terms (assuming $x$ is nonnegative.) For example,

$$\sqrt\frac{x^3}{50} = \frac{\sqrt{x^3}}{\sqrt{50}}$$

Then, you want to try to rationalize the denominator OR try to manipulate the denominator as I did above to make it more friendly (i.e. $\sqrt{25}\cdot{\sqrt{2}} = \sqrt{50}$, and since you know that $\sqrt{25} = 5$, $\sqrt{50}$ must equal $5\sqrt{2}$.

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This answer seems needlessly complicated and incomplete. Notice $\sqrt{x^3} = \sqrt{x^2}\sqrt{x} = x\sqrt{x}$. Fractional exponents are best for certain purposes, but when you're talking about "simplest radical form", this is the way to do it. – Michael Hardy Jun 13 '12 at 21:28
...also, for some purposes one rationalizes the denominator, so you'd change $\dfrac{1}{2\sqrt{3}}$ to $\dfrac{\sqrt{3}}{6}$. – Michael Hardy Jun 13 '12 at 21:30
Which leaves a question: Why should one rationalize the denominator? If students fail to ask that question, I might suspect them of being there for the purpose of putting the required course behind them so they can forget it. I wonder if anyone's asked that question here? – Michael Hardy Jun 13 '12 at 21:31
That question was asked here:… – Michael Hardy Jun 13 '12 at 21:42
@MichaelHardy Thanks for the comments. Looking back, I agree that I probably should have written it as $x\sqrt{x}$ on top. I remember being taught that it was fine to leave square root terms in the denominator. One of my high school math teachers stated the only reason it was the norm of society was because it was easier? to use for typewriters since supposedly people had a difficult time (perhaps it was impossible on the original typewriters?) of putting a square root in the bottom. Note that I never questioned this; perhaps I need to look into it more. To each his own, I suppose. – Joe Jun 13 '12 at 22:28

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