# basic surds

• February 22nd 2011, 07:50 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?
• February 22nd 2011, 07:59 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?

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.
• February 22nd 2011, 08:08 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.
• February 22nd 2011, 08:39 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?
• February 23rd 2011, 03:16 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.