# Thread: Rationalizing radicals in the denominator

1. ## Rationalizing radicals in the denominator

It's little difficult for me...

$\frac{2}{\sqrt[5]{4}}$

2. $\frac{2}{\sqrt[5]{4}} = \frac{2}{4^{(1/5)}}= \frac{2}{(2)^{2(1/5)}} = \frac{2}{2^{(2/5)}}$

NOw complete it

3. I don't understand. Correct answer is $\sqrt[5]{8}$

4. YOu need to look at your books/notes:

$\frac{2}{\sqrt[5]{4}} = \frac{2}{4^{(1/5)}}= \frac{2}{(2)^{2(1/5)}} = \frac{2}{2^{(2/5)}} = 2^{1-\frac{2}{5}} = 2^{(3/5)}=2^{3(1/5)} = 8^{(1/5)} = \sqrt[5]{8}$

Clear??

5. $\dfrac{2}{{\sqrt[5]{4}}} = \dfrac{2}{{\sqrt[5]{{2^2 }}}}\dfrac{{\sqrt[5]{{2^3 }}}}{{\sqrt[5]{{2^3 }}}} = \sqrt[5]{8}$

6. I see. I just think you leave answer in that first post. I miss to read complete it. Sorry. I'm absent-minded today.

7. Originally Posted by Lil
It's little difficult for me...

$\frac{2}{\sqrt[5]{4}}$
Express the numerator as a fifth-root also...

$\displaystyle\frac{2}{\sqrt[5]4}}=\frac{\sqrt[5]{2^5}}{\sqrt[5]4}=\sqrt[5]{\frac{2^5}{2^2}}$

8. I solve this now. It's correct or no?
$\frac{4}{3\sqrt{5}-4}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5}-4){(3\sqrt{5}+4)}}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5 })^2-(4)^{2}}=\frac{4(3\sqrt{5}+4)}{29}.$

9. Yes!

It can also be expressed as:

$\frac{4(3\sqrt{5}+4)}{29} = \frac{12\sqrt{5}}{29} + \frac{16}{29}$

10. Originally Posted by Lil
I solve this now. It's correct or no?

$\frac{4}{3\sqrt{5}-4}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5}-4){(3\sqrt{5}+4)}}=\frac{4(3\sqrt{5}+4)}{(3\sqrt{5 }-4)^{2}}=\frac{4(3\sqrt{5}+4)}{29}.$
You have the right idea (surd conjugate) and your final answer has rationalised the denominator and is correct.

However $(3\sqrt{5}-4)(3\sqrt{5}+4)=(3\sqrt{5})^2-4^2=29$

$(3\sqrt{5}-4)(3\sqrt{5}+4)\ \ne\ (3\sqrt{5}-4)^2$

11. Originally Posted by Archie Meade
You have the right idea (surd conjugate) and your final answer has rationalised the denominator and is correct.

However $(3\sqrt{5}-4)(3\sqrt{5}+4)=(3\sqrt{5})^2-4^2=29$

$(3\sqrt{5}-4)(3\sqrt{5}+4)\ \ne\ (3\sqrt{5}-4)^2$
What a stupid mistake!!! I know this rule. My head don't work today. Thanks.

12. I won't make new topic. Just continue this, because my work is with roots.
Stuck with this.
Just calculate expression:

$2*\left(\frac{2}{\sqrt{6}+3} +\frac{3}{\sqrt{6}-2}-\frac{5}{\sqrt{6}}\right)$

13. $2*\left(\frac{2}{\sqrt{6}+3} +\frac{3}{\sqrt{6}-2}-\frac{5}{\sqrt{6}}\right)$

$= 2*\left( \left[\frac{2}{\sqrt{6}+3} \times \frac{\sqrt{6}-3}{\sqrt{6}-3}\right]+ \left[\frac{3}{\sqrt{6}-2} \times \frac{\sqrt{6}+2}{\sqrt{6}+2}\right]-\left[\frac{5}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}\right]\right)$

Now Simplify.....

14. I won't write all solution... It's clear now. Answer I get 10. I think it's correct.

15. Yes!!

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