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How would i simplify fully the following expression?

$\dfrac{{\sqrt 2}({x^3})}{\sqrt{\frac {32}{x^2}}}$

So far i have got this

$\dfrac{{\sqrt 2}{x^3}}{{\frac{\sqrt 32}{\sqrt x^2}}}$ = $\dfrac{{\sqrt 2}{x^3}}{{\frac{4\sqrt 2}{x}}}$

Am not quite sure if this is correct however, could someone help explain how i would simplify this expression?

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5 Answers 5

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There is a mistake in the OP. Recall that $\sqrt{x^2}=|x|\ne x$ when $x<0$. To simplify, we can write

$$\frac{\sqrt 2 x^3}{\sqrt{\frac{32}x}}=\frac{\sqrt 2 x^3}{\frac{4\sqrt 2}{|x|}}=\frac{x^3|x|}{4}$$

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  • $\begingroup$ Am not quite sure what |x| is supposed to mean? $\endgroup$
    – Zochonis
    Commented Oct 1, 2016 at 18:21
  • $\begingroup$ That is the absolute value sign. $\endgroup$
    – Mark Viola
    Commented Oct 1, 2016 at 18:37
  • $\begingroup$ Oh yes, thanks a bunch! $\endgroup$
    – Zochonis
    Commented Oct 1, 2016 at 18:49
  • $\begingroup$ You're welcome. My pleasure. -Mark $\endgroup$
    – Mark Viola
    Commented Oct 1, 2016 at 19:02
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There isn't a universal notion of when an expression is “fully simplified,” but you generally want to have a simple fraction (not a complex one, where numerator or denominator are themselves functions). In this expression I would also try to combine all the square roots into a single radical, and put that radical in the numerator.

First, make the fraction simple: $$ \frac{\sqrt{2} x^3}{\sqrt{32/x^2}} = \frac{\sqrt{2} x^3}{\sqrt{32}/\sqrt{x^2}} = \frac{\sqrt{2}\sqrt{x^2} x^3}{\sqrt{32}} $$ You are probably expected to simplify $\sqrt{x^2} = x$, although this is only true when $x\geq0$. You might want to ask your teacher if you are supposed to assume $x$ is positive.

Combining the $\sqrt{2}$ in the numerator with $\sqrt{32}$ in the denominator gives $$ \frac{x^4}{\sqrt{16}} = \frac{x^4}{4} $$

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Note that $\sqrt{x^2} = |x|$

The simplified expression would be $\frac{x^3|x|}{4}$

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It is correct so far, continue by multiplying the reciprocal of the fraction and simplifying the answer.

x^4/4 if x is positive and - x^4/4 if x is negative

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    $\begingroup$ Wrong, because $\sqrt{x^2}=|x|$. $\endgroup$ Commented Oct 1, 2016 at 17:34
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$\frac{\sqrt{2}x^3}{\sqrt{\frac{32}{x^2}}} = \frac{\sqrt{2}x^3}{\frac{\sqrt{32}}{\sqrt{x^2}}} = \frac{\sqrt{2}x^3}{\frac{4\sqrt{2}}{x}} = \frac{x^3}{\frac{4}{x}} = \frac{x^4}{4}$

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  • $\begingroup$ Wrong, because $\sqrt{x^2}=|x|$. $\endgroup$ Commented Oct 1, 2016 at 17:34

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