# 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

and got an answer of

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 ...

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

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

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

HI, thanks for replying,

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