Why is $\frac{\sqrt{32}}{\sqrt{14D}}=\frac{4\sqrt{7D}}{7D}$ and not $2\sqrt{8}$?
I am asked to simplify $\frac{\sqrt{32}}{\sqrt{14D}}$ and am provided with the solution $\frac{4\sqrt{7D}}{7D}$.
(Side question, in the solution why can't $\sqrt{7D}$ and $7D$ cancel out since one is in he denominator so if multiplying out would it not be $\sqrt{7D} * 7D = 0$?)
I gave this question a try but arrived at $2\sqrt{8}$. Here's my working:
(multiply out the radicals in the denominator)
$\frac{\sqrt{32}}{\sqrt{14D}}$ =
$\frac{\sqrt{32}}{\sqrt{14D}} * \frac{\sqrt{14D}}{\sqrt{14D}}$ =
$\frac{\sqrt{32}\sqrt{14D}}{14D}$ =
(Use product rule to split out 32 in numerator)
$\frac{\sqrt{4}\sqrt{8}\sqrt{14D}}{14D}$ =
$\frac{2\sqrt{8}\sqrt{14D}}{14D}$ =
Then, using my presumably flawed logic in my side question above I cancelled out $14D$ to arrive at $2\sqrt{8}$
Where did I go wrong and how can I arrive at $\frac{4\sqrt{7D}}{7D}$?
$\endgroup$ 84 Answers
$\begingroup$Hint: Multiplying numerator and denomninator by $$\sqrt{14D}$$ we get $$\frac{\sqrt{32}\sqrt{14D}}{14D}$$ and this is equal b$$\frac{4\sqrt{2}{\sqrt{2}}\sqrt{7D}}{14D}$$ and this is equal to $$\frac{4\sqrt{7D}}{7D}$$ for $$D\neq 0$$$$\sqrt{32}=\sqrt{2\cdot 16}=4\sqrt{2}$$ and $$\sqrt{14}=\sqrt{2}\sqrt{7}$$
$\endgroup$ 2 $\begingroup$Your paragraph with side note shows two major misconceptions.
The first is that $\sqrt{7D}$ is not the same as $7D$ so you can't cancel them. As long as $x \gt 0$ so things make sense $\frac {\sqrt x}x=\frac {x^{1/2}}x=\frac 1{x^{1/2}}$ by the laws of exponents.
Second, if you could cancel them you would be claiming that $\frac {\sqrt{7D}}{7D}=1$. Their product would definitely not be $0$. Saying you are canceling things from a fraction is a good way to make mistakes. I find it much better to think in terms of multiplying the fraction by $1$ in either the form $\frac xx$ or the form $\dfrac {\frac 1x}{\frac 1x}$. When you have terms added in the numerator or denominator that prompts you to distribute the new factor properly and also alerts you to the fact that if you have nothing left it should be $1$, not $0$.
$\endgroup$ 1 $\begingroup$I’m not exactly sure why you thought $\frac{\sqrt{14D}}{14D}$ can be cancelled. The square root of a value is not equal to itself unless the value is $1$, so that’s clearly false. Usually, when you’re unsure about simplifications, you can try inserting a value to check if it is valid. Instead, you can simplify as follows:
$$\frac{\sqrt{32}}{\sqrt{14D}} = \frac{\sqrt{2^5}}{\sqrt{14D}} = \frac{\sqrt{\left(2^2\right)^2\cdot 2}}{\sqrt{2\cdot 7D}} = \frac{2^2\sqrt{2}}{\sqrt{2}\sqrt{7D}} = \frac{2^2}{\sqrt{7D}} = \frac{4}{\sqrt{7D}}$$
Rationalization yields
$$\frac{4}{\sqrt{7D}}\cdot\frac{\sqrt{7D}}{\sqrt{7D}} = \frac{4\sqrt{7D}}{7D}$$
$\endgroup$ $\begingroup$$\frac{\sqrt{32}}{\sqrt{14D}}=\frac{4\sqrt{2}}{\sqrt{2}\sqrt{7D}}=\frac{4}{\sqrt{7D}}=\frac{4}{\sqrt{7D}}\frac{\sqrt{7D}}{\sqrt{7D}}=\frac{4\sqrt{7D}}{7D}$
but $\frac{4}{\sqrt{7D}}=2\sqrt{8}$ then $\sqrt{14D}=1$ so $D=\frac{1}{14}$.
$\endgroup$
