![Why is $\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ $15\sqrt{5}$ and not $15\sqrt[4]{25}$?](https://liberty.io/uploads/why-is-frac-15-sqrt-4-125-sqrt-4-5-15-sqrt-5-and-not-15-sqrt-4-25_720.jpg)
$\begingroup$I have an expression I am to simplify:
$$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$$
I arrived at $15\sqrt[4]{25}$. My textbook tells me that the answer is in fact $15\sqrt{5}$. Here is my thought process:
$\frac{15\sqrt[4]{125}}{\sqrt[4]{5}}$ = $\frac{15*\sqrt[4]{5}*\sqrt[4]{25}}{\sqrt[4]{5}}$ =
(cancel out $\sqrt[4]{5}$ present in both numerator and denominator) leaving: $$15\sqrt[4]{25}$$
Where did I go wrong and how can I arrive at $15\sqrt{5}$?
$\endgroup$13 Answers
$\begingroup$Write $$\sqrt[4]{\frac{125}{5}}=\sqrt[4]{25}=\sqrt{5}$$
$\endgroup$$\begingroup$It turns out that $\sqrt[4]{25}=\sqrt{5}$. This is because $25=5^2$, so that $\sqrt[4]{25}=\sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$ So, you are correct, as is the book.
$\endgroup$1$\begingroup$$$\frac{\sqrt[4] {125}}{\sqrt[4] 5} = \frac{\sqrt[4] {5^3}}{\sqrt[4] 5} = \sqrt[4]{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$
Even quicker, $\sqrt[4]{25}$ means $\sqrt{\sqrt{25}}$, which becomes $\sqrt{5}$.
$\endgroup$
