# basic surds

• Feb 22nd 2011, 08:50 PM
harun
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?

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

The only way I could think of solving it was equating it as $3=\sqrt{a}$
and $2\sqrt{5}=b$ and then solving for a and b.
But that seem highly dubious.
Can anyone tell me if this method is correct?
Thank you in advance.
• Feb 22nd 2011, 08:59 PM
TheEmptySet
Quote:

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?

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

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

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

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

Now for any $a > 0$ just pick an $a$ and plug into the equation to find another b. You can graph the function as well to see the set of solutions.
• Feb 22nd 2011, 09:08 PM
pickslides
Something different to TES.

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

Now finish it off.
• Feb 22nd 2011, 09:39 PM
harun
Quote:

Originally Posted by pickslides
Something different to TES.

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

Now finish it off.

Thanks for the replies, I think this might be the solution. So, this would be

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

$

Or a=45 and b=10. Is this correct?
• Feb 23rd 2011, 04:16 AM
HallsofIvy
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.