## Simplifying a radical expression with all variables

Simplificatation, evaluation, linear equations, linear graphs, linear inequalities, basic word problems, etc.
maroonblazer
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### Simplifying a radical expression with all variables

Hi,
I have the following problem:

$\sqrt[n]{x^{n+2}} * \sqrt[n]{x^{n-2}}$, where n is an integer >2 and variables in the radicand of an even index are non-negative.

I'm able to reduce/simplify the left side of the multiplication to $x^2$, however I'm struggling with the right-hand side.

Ideas?

stapel_eliz
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maroonblazer wrote:I have the following problem:

$\sqrt[n]{x^{n+2}} * \sqrt[n]{x^{n-2}}$, where n is an integer >2 and variables in the radicand of an even index are non-negative.

You refer to "sides", but that is a term used for equations, where there is an "equals" sign and then two sides. However, what you have posted is an expression; that is, there is no "equals" sign.

Are you perhaps referring to the first of the two factors? If so, how are you "reducing" this to $x^2$?

maroonblazer
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Joined: Thu Aug 12, 2010 11:16 am
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### Re: Simplifying a radical expression with all variables

Ok, holidays are over, back at it...

Thanks for the tip. Here's what I'm doing:

$\sqrt[n]{(x^{n+2})(x^{n-2})}$

Now I start to get shaky. I add the exponents, right? Which would give:

$\sqrt[n]{(x^n)(x^n)}$ (since the +2 and -2, when summed, cancel each other out).

Since $\sqrt[n]{x^n^} = \sqrt{x}$ then the expression above simplifies to:

$\sqrt{x^2}$ which simplifies further to:

$x^2$

Do I have that right?

stapel_eliz
Posts: 1738
Joined: Mon Dec 08, 2008 4:22 pm
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maroonblazer wrote:$\sqrt[n]{(x^{n+2})(x^{n-2})}$

Now I start to get shaky. I add the exponents, right? Which would give:

$\sqrt[n]{(x^n)(x^n)}$

Actually, the sum of $n\, +\, 2$ and $n\, -\, 2$ is $2n$, so the simplification should be:

. . . . .$\sqrt[n]{(x^{n+2})(x^{n-2})}\, =\, \sqrt[n]{x^{2n}}\, =\, \sqrt[n]{\left(x^2\right)^n}$

maroonblazer wrote:Since $\sqrt[n]{x^n^} = \sqrt{x}$ then....

Actually:

. . . . .$\sqrt[n]{x^n}\, =\, \left{\begin{array}{rl}x,&\mbox{for }\, n\, \mbox{ odd}\\|x|,&\mbox{for }\, n\, \mbox{ even}\end{array}$

...so the previous expression simplifies as $|x^2|$. However, since the square is never negative, the absolute-value bars aren't needed, and you end up with your final result.

I make the big deal about the absolute-value bars because you could be given a "simplification" exercise along these lines where the parity (oddness or evenness) of the exponent and / or the radical will matter. Keep an eye out on your next test; it's often a trick question on exams.