# Simplifying radicals: 2 * 5(sqrt) / 5(sqrt) - 3

#### skaterchicky23

##### New member
I have a question of 2 * 5(sqrt)
-------------
5(sqrt)-3

I took this problem and took the following steps:

2*5(sqrt) 5(sqrt)+3
---------- * -----------
5(sqrt)-3 5(sqrt)+3

2(5+3)
------
(-3)^2+(5(sqrt)^2

=16 8
------- or ---
34 17

is this correct?

I have tried looking up online for what else I would do instead, but the examples are showing similar to this.

#### skeeter

##### Senior Member

if you mean ...

$$\displaystyle \frac{2 \sqrt{5}}{\sqrt{5} - 3}$$

then ...

$$\displaystyle \frac{2 \sqrt{5}}{\sqrt{5} - 3} \cdot \frac{\sqrt{5} + 3}{\sqrt{5} + 3} =$$

$$\displaystyle \frac{10 + 6\sqrt{5}}{5 - 9} =$$

$$\displaystyle - \frac{5 + 3\sqrt{5}}{2}$$

#### skaterchicky23

##### New member

skeeter said:
if you mean ...

$$\displaystyle \frac{2 \sqrt{5}}{\sqrt{5} - 3}$$

then ...

$$\displaystyle \frac{2 \sqrt{5}}{\sqrt{5} - 3} \cdot \frac{\sqrt{5} + 3}{\sqrt{5} + 3} =$$

$$\displaystyle \frac{10 + 6\sqrt{5}}{5 - 9} =$$

$$\displaystyle - \frac{5 + 3\sqrt{5}}{2}$$

How do you get the 10 in the 3rd equation?
I'm confused by how the 10 became part of the equation, if you could explain that would be great

#### skeeter

##### Senior Member
$$\displaystyle 2\sqrt{5} \cdot \sqrt{5} = 2(\sqrt{5} \cdot \sqrt{5}) = 2(5) = 10$$