First if you had to rationalise 2/radical: 5 why does it become 2 and radical 5 over 5. Why is not 10/25?

Second question is 4/3 and radical 7, why do we multiply 3 and radical 7 to get 21?

I hope i've explained it well...

- Thread starter Luke55
- Start date

First if you had to rationalise 2/radical: 5 why does it become 2 and radical 5 over 5. Why is not 10/25?

Second question is 4/3 and radical 7, why do we multiply 3 and radical 7 to get 21?

I hope i've explained it well...

- Joined
- Jan 27, 2012

- Messages
- 6,636

The basic idea is that \(\displaystyle \sqrt{a}(\sqrt{a})= (\sqrt{a})^2= a\). So to "get rid of" a square rootHi, i have just a few questions on how Rationalizing the Denominator works.

First if you had to rationalise 2/radical: 5 why does it become 2 and radical 5 over 5. Why is not 10/25?

you multiply by that same square root, Also I hope you understand that, in order not to change the actual value of a fraction you have to multiply both numerator and denominator by the same thing.

To "get rid of" \(\displaystyle \sqrt{5}\) in the denominator, multiply

I am puzzled as to why you would think it should be \(\displaystyle \frac{10}{25}\)! For the numerator to go from 2 to 10 you must have multiplied by 5. But multiplying the denominator by 5 does not give 25, it gives \(\displaystyle 5\sqrt{5}\). In fact, it should be easy to see that \(\displaystyle \frac{10}{25}= \frac{2\cdot 5}{5\cdot 5}= \frac{2}{5}\), not \(\displaystyle \frac{2}{\sqrt{5}}\). Or are you simply not clear on what "\(\displaystyle \sqrt{}\)" means?

I think you mean \(\displaystyle \frac{4}{3\sqrt{7}}\)?Second question is 4/3 and radical 7,

Wewhy do we multiply 3 and radical 7 to get 21?

\(\displaystyle \frac{4}{3\sqrt{7}}\frac{\sqrt{7}}{\sqrt{7}}= \frac{4\sqrt{7}}{3\sqrt{7}\cdot \sqrt{7}}= \frac{4\sqrt{7}}{3(\sqrt{7})^2}= \frac{4\sqrt{7}}{3\cdot 7}= \frac{4\sqrt{7}}{21}\)

I hope i've explained it well...

The square root of 5 is in the neighborhood of 2.2. Since \(\displaystyle 5 \ne \sqrt{5}\), it is likewise true that \(\displaystyle 5 * 5 \ne 5 * \sqrt{5}\).First if you had to rationalise 2/radical: 5 why does it become 2 and radical 5 over 5. Why is not 10/25?

You can use a fraction that is the same on the top and the bottom (which is equal to 1) and multiply with it to move terms around within another fraction. Consider the following:

\(\displaystyle \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} * 1\)

\(\displaystyle \frac{2}{\sqrt{5}} = \frac{2}{\sqrt{5}} * \frac{\sqrt{5}}{\sqrt{5}}\)

\(\displaystyle \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{\sqrt{5}^2}\)

\(\displaystyle \frac{2}{\sqrt{5}} = \frac{2 \sqrt{5}}{5}\)

- Joined
- Jan 27, 2012

- Messages
- 6,636

I'm not sure what you mean by this. It's pretty straight forward. \(\displaystyle \sqrt{a}\) is the positive number, x, such \(\displaystyle x^2= a\). The square root of 4 is 2 because \(\displaystyle 2^2= 2(2)= 4\). The square root or 9 is 3 because \(\displaystyle 3^2= 3(3)= 9\). The square root of 2.25 is 1.5 because \(\displaystyle 1.5^2= 1.5(1.5)= 2.25\).I don't understand the square root sign from the looks of it.

Last edited by a moderator: