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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?
There is a mistake in the OP. Recall that $\sqrt{x^2}=|x|\ne x$ when $x<0$. To simplify, we can write
$$\frac{\sqrt 2 x^3}{\sqrt{\frac{32}x}}=\frac{\sqrt 2 x^3}{\frac{4\sqrt 2}{|x|}}=\frac{x^3|x|}{4}$$
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 $$ \frac{x^4}{\sqrt{16}} = \frac{x^4}{4} $$
Note that $\sqrt{x^2} = |x|$
The simplified expression would be $\frac{x^3|x|}{4}$
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
$\frac{\sqrt{2}x^3}{\sqrt{\frac{32}{x^2}}} = \frac{\sqrt{2}x^3}{\frac{\sqrt{32}}{\sqrt{x^2}}} = \frac{\sqrt{2}x^3}{\frac{4\sqrt{2}}{x}} = \frac{x^3}{\frac{4}{x}} = \frac{x^4}{4}$
