# Why is $\frac{15\sqrt{125}}{\sqrt{5}}$ $15\sqrt{5}$ and not $15\sqrt{25}$?

I have an expression I am to simplify:

$$\frac{15\sqrt{125}}{\sqrt{5}}$$

I arrived at $$15\sqrt{25}$$. My textbook tells me that the answer is in fact $$15\sqrt{5}$$. Here is my thought process:

$$\frac{15\sqrt{125}}{\sqrt{5}}$$ = $$\frac{15*\sqrt{5}*\sqrt{25}}{\sqrt{5}}$$ =

(cancel out $$\sqrt{5}$$ present in both numerator and denominator) leaving: $$15\sqrt{25}$$

Where did I go wrong and how can I arrive at $$15\sqrt{5}$$?

• The fourth root is the square root of the square root ... – Ethan Bolker Jan 3 '19 at 16:54

$$\frac{\sqrt {125}}{\sqrt 5} = \frac{\sqrt {5^3}}{\sqrt 5} = \sqrt{5^2} = 5^{\frac{2}{4}} = 5^{\frac{1}{2}} = \sqrt 5$$
Even quicker, $$\sqrt{25}$$ means $$\sqrt{\sqrt{25}}$$, which becomes $$\sqrt{5}$$.
Write $$\sqrt{\frac{125}{5}}=\sqrt{25}=\sqrt{5}$$
It turns out that $$\sqrt{25}=\sqrt{5}$$. This is because $$25=5^2$$, so that $$\sqrt{25}=\sqrt{5^2}=(5^2)^{1/4}=5^{1/2}=\sqrt{5}.$$ So, you are correct, as is the book.