1. ## basic surds

I was helping someone with this problem and I thought you could only simplify it and nothing more. Is it possible to find values of a and b?

$\displaystyle \frac{3}{2\sqrt{5}} = \frac{\sqrt{a}}{b}$

The only way I could think of solving it was equating it as $\displaystyle 3=\sqrt{a}$
and $\displaystyle 2\sqrt{5}=b$ and then solving for a and b.
But that seem highly dubious.
Can anyone tell me if this method is correct?

2. Originally Posted by harun
I was helping someone with this problem and I thought you could only simplify it and nothing more. Is it possible to find values of a and b?

$\displaystyle \frac{3}{2\sqrt{5}} = \frac{\sqrt{a}}{b}$

The only way I could think of solving it was equating it as $\displaystyle 3=\sqrt{a}$
and $\displaystyle 2\sqrt{5}=b$ and then solving for a and b.
But that seem highly dubious.
Can anyone tell me if this method is correct?

Well that is one solution but this equation has an infinite number of solutions

$\displaystyle \displaystyle \frac{3}{2\sqrt{5}} = \frac{\sqrt{a}}{b} \iff b=\frac{2\sqrt{5a}}{3}$

Now for any $\displaystyle a > 0$ just pick an $\displaystyle a$ and plug into the equation to find another b. You can graph the function as well to see the set of solutions.

3. Something different to TES.

$\displaystyle \displaystyle \frac{3}{2\sqrt{5}} = \frac{3\sqrt{5}}{2\times \sqrt{5}\times \sqrt{5}} = \frac{3\sqrt{5}}{10}= \frac{\sqrt{9}\sqrt{5}}{10}$

Now finish it off.

4. Originally Posted by pickslides
Something different to TES.

$\displaystyle \displaystyle \frac{3}{2\sqrt{5}} = \frac{3\sqrt{5}}{2\times \sqrt{5}\times \sqrt{5}} = \frac{3\sqrt{5}}{10}= \frac{\sqrt{9}\sqrt{5}}{10}$

Now finish it off.
Thanks for the replies, I think this might be the solution. So, this would be

$\displaystyle \frac{\sqrt{9}\sqrt{5}}{10} = \frac{\sqrt{45}}{10} = \frac{\sqrt{a}}{b}$

Or a=45 and b=10. Is this correct?

5. Yes, that is correct. That is referred to as "rationalizing the denominator". Note that in your original post you did not say that a and b must be integers but everyone assumed that was what you meant.