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Simplification Problem

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anemone

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Feb 14, 2012
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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

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Feb 24, 2012
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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:

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

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Feb 14, 2012
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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 $-u^5+5u=0$.:eek: Hehehe...
 
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Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
8,780
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

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Feb 24, 2012
13,775
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.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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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!
 

Klaas van Aarsen

MHB Seeker
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Mar 5, 2012
8,780
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

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Feb 24, 2012
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(Rofl) I am the typo king! (Tmi)
 

Chris L T521

Well-known member
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Jan 26, 2012
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(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

Administrator
Staff member
Feb 24, 2012
13,775
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)