MHB Can this Simplification Problem be Solved Using Elementary Methods?

[anemone]

Problem:
Simplify $ \dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$.

I just can't see a way to solve it...

I hope someone could give me some hints if this problem could be solved using only elementary methods.

Thanks in advance.

 

[MarkFL]

Let's let:

$\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=x$

Let also:

$u=\sqrt[4]{5}$

and we have:

$\dfrac{2}{\sqrt{4-3u+2u^2-u^3}}=x$

$2=x\sqrt{4-3u+2u^2-u^3}$

$4=x^2(4-3u+2u^2-u^3)$

Now, if we observe that:

$-u^5+5u+4=(1+u)^2(4-3u+2u^2-u^3)$

Then we may state that:

$\displaystyle x=\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}$

 

[anemone]

Let's let:

$\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=x$

Let also:

$u=\sqrt[4]{5}$

and we have:

$\dfrac{2}{\sqrt{4-3u+2u^2-u^3}}=x$

$2=x\sqrt{4-3u+2u^2-u^3}$

$4=x^2(4-3u+2u^2-u^3)$

Now, if we observe that:

$-u^5+5u+4=(1+u)^2(4-3u+2u^2-u^3)$

Then we may state that:

$\displaystyle x=\dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}$

Bravo and thanks, Mark!(Smile) Admittedly, I would have to think for a short while before I realized $-u^5+5u=0$. Hehehe...

 

[I like Serena]

Hey anemone! :)

Problem:
Simplify $ \dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$.

I just can't see a way to solve it...

I hope someone could give me some hints if this problem could be solved using only elementary methods.

Thanks in advance.

To simplify such an expression, the trick is to multiply numerator and denominator by something smart.
Now what kind of smart thing can we come up with?
Let's pick $(1+\sqrt[4]{5})$ or if it does not work we can next try $(1-\sqrt[4]{5})$.

That brings:

$\dfrac{2(1+\sqrt[4]{5})}{\sqrt{(1+\sqrt[4]{5})^2(4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125})}}$

It's a bit of work to work that out, but then we'll get:

$\dfrac{2(1+\sqrt[4]{5})}{\sqrt{4 + 5\sqrt[4]{5}-(\sqrt[4]{5})^5}} = \dfrac{2(1+\sqrt[4]{5})}{\sqrt{4 + 5\sqrt[4]{5}-5\sqrt[4]{5}}} = 1+\sqrt[4]{5}$@Mark: Btw, can you replace your $-u^5 + 5u^4 + 4$ by $-u^5 + 5u + 4$, since otherwise it won't come out as 4?

 

[MarkFL]

I should add, that the "observation" I spoke of might come about by using division to find:

$\displaystyle \frac{-u^5+5u+4}{4-3u+2u^2-u^3}=(1+u)^2$

I like Serena is correct (thank you), I did have a typo in my first post, which I will correct now.

 

[anemone]

Hey anemone! :)
To simplify such an expression, the trick is to multiply numerator and denominator by something smart.
Now what kind of smart thing can we come up with?
Let's pick $(1+\sqrt[4]{5})$ or if it does not work we can next try $(1-\sqrt[4]{5})$.

Thanks, ILikeSerena...that is an useful and handy hint to me!

 

[I like Serena]

I should add, that the "observation" I spoke of might come about by using division to find:

$\displaystyle \frac{-u^5+5u+4}{4-3u+2u^2-u^3}=(1+u)^2$

ILikeSerena is correct (thank you), I did have a type in my first post, which I will correct now.

Thanks, ILikeSerena...that is an useful and handy hint to me!

It must be nice that you can fix typos in quotes.
I'm always afraid to make typos, since if someone quotes me I cannot correct it anymore.
And hey, you two just made another typo! ;)

 

[MarkFL]

(Rofl) I am the typo king! (Tmi)

 

[Chris L T521]

(Rofl) I am the typo king! (Tmi)

I would have been more amused if you made a type in that declaration (that was intentional, by the way). XD

 

[MarkFL]

I would have been more amused if you made a type in that declaration (that was intentional, by the way). XD

I did miss a golden opportunity there! (Smirk)