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Simplification Challenge

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  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
Simplify $\large \dfrac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}$.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
My solution:

Let \(\displaystyle x=\sqrt[4]{5}\)

I would try to see if there is a solution for $a$ in:

\(\displaystyle \left(1+ax \right)^2\left(4-3x+2x^2-x^3 \right)=4\)

\(\displaystyle -a^2x^5+(2a^2-2a)x^4+(-3a^2+4a-1)x^3+(4a^2-6a+2)x^2+(8a-3)x+4=4\)

Now, we have:

\(\displaystyle x^5=5x\) and \(\displaystyle x^4=5\) hence we may write (after factoring):

\(\displaystyle (a-1)(1-3a)x^3+2(a-1)(2a-1)x^2+(a-1)(3-5a)x+10a(a-1)=0\)

Thus, we find $a=1$. And so we may write:

\(\displaystyle \frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}\)
 
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  • #3

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
My solution:

Let \(\displaystyle x=\sqrt[4]{5}\)

I would try to see if there is a solution for $a$ in:

\(\displaystyle \left(1+ax \right)^2\left(4-3x+2x^2-x^3 \right)=4\)

\(\displaystyle -a^2x^5+(2a^2-2a)x^4+(-3a^2+4a-1)x^3+(4a^2-6a+2)x^2+(8a-3)x+4=4\)

Now, we have:

\(\displaystyle x^5=5x\) and \(\displaystyle x^4=5\) hence we may write (after factoring):

\(\displaystyle (a-1)(1-3a)x^3+2(a-1)(2a-1)x^2+(a-1)(3-5a)x+10a(a-1)=0\)

Thus, we find $a=1$. And so we may write:

\(\displaystyle \frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}=1+\sqrt[4]{5}\)
Thanks for participating, MarkFL! You're very smart in checking if your first trial over the equality holds and if it does, what value of $a$ does it takes. Well done!