# Simplifying help

• Nov 24th 2012, 09:07 AM
gfbrd
Simplifying help
Hey there I need some help simplifying this problem
hope you guys can see it

Attachment 25901

oh and that is an 8th root if you guys are wondering
also I'm looking for the steps to go about simplifying this problem not just the answer thanks
• Nov 24th 2012, 09:17 AM
skeeter
Re: Simplifying help
$\displaystyle \sqrt[8]{\frac{2207-987\sqrt{5}}{2}}$

what makes you think the expression can be simplified further than the form you've written?
• Nov 24th 2012, 09:19 AM
gfbrd
Re: Simplifying help
because i know the simplified form

i just dont know how to go about simplifying it thats all
• Nov 24th 2012, 09:30 AM
skeeter
Re: Simplifying help
care to share that "simplified form" ... because my CAS can't do it.
• Nov 24th 2012, 09:36 AM
gfbrd
Re: Simplifying help
it should be
(3 + sqrt(5))/ 2
• Nov 24th 2012, 09:45 AM
skeeter
Re: Simplifying help
uhh ...

$\displaystyle \sqrt[8]{\frac{2207-987\sqrt{5}}{2}} \approx 0.381966010851...$

$\displaystyle \frac{3+\sqrt{5}}{2} \approx 2.61803398875...$

???
• Nov 24th 2012, 09:56 AM
gfbrd
Re: Simplifying help
oh sorry i made a mistake in typing the problem
it should be + 987sqrt(5) not subtract
• Nov 24th 2012, 10:14 AM
skeeter
Re: Simplifying help
That looks better to the calculator, but I can't see a quick simplification at this time. If it doesn't come to me, hopefully someone else can help.
• Nov 24th 2012, 10:25 AM
gfbrd
Re: Simplifying help
oh alright thanks for helping
• Nov 24th 2012, 05:10 PM
Prove It
Re: Simplifying help
Quote:

Originally Posted by gfbrd
Hey there I need some help simplifying this problem
hope you guys can see it

Attachment 25901

oh and that is an 8th root if you guys are wondering
also I'm looking for the steps to go about simplifying this problem not just the answer thanks

A word of advice, ALL "roots" of fractions are considered unsimplified. You are expected to always rationalise denominators at the very least.

\displaystyle \displaystyle \begin{align*} \sqrt[8]{\frac{2207 + 987\sqrt{5}}{2}} &= \frac{\sqrt[8]{2207 + 987\sqrt{5}}}{\sqrt[8]{2}} \\ &= \frac{\sqrt[8]{2207 + 987\sqrt{5}}\cdot \left( \sqrt[8]{2} \right)^7}{\sqrt[8]{2} \cdot \left(\sqrt[8]{2}\right)^7} \\ &= \frac{\sqrt[8]{2207 + 987\sqrt{5}} \cdot \sqrt[8]{2^7}}{2} \end{align*}

I'll have to think about how the top simplifies to \displaystyle \displaystyle \begin{align*} 3 + \sqrt{5} \end{align*}...
• Nov 24th 2012, 07:02 PM
Soroban
Re: Simplifying help
Hello, gfbrd!

Quote:

$\displaystyle \text{It should be }\:\frac{3 + \sqrt{5}}{2}$ . . . . no

The original expression has a "minus": .$\displaystyle \sqrt[8]{\frac{2207 \:{\color{red}-}\: 987\sqrt{5}}{2}}$

You must mean: .$\displaystyle \frac{3\:{\color{red}-}\:\sqrt{5}}{2}$
. . which happens to equal: .$\displaystyle \frac{1}{\phi^2}$
. . where $\displaystyle \phi \,=\,\tfrac{1+\sqrt{5}}{2}$, the Golden Ratio.

It can be shown that: .$\displaystyle \frac{1}{\phi^{16}} \:=\:\frac{2207 - 987\sqrt{5}}{2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I'd like to share a "trick" for raising $\displaystyle \phi$ to higher powers.

One of the root of the quadratic equation: .$\displaystyle x^2 - x - 1 \:=\:0$ .is $\displaystyle \phi.$

This means: .$\displaystyle \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}$

Divide by $\displaystyle \phi\!:\;\phi \:=\:1 + \frac{1}{\phi} \quad\Rightarrow\quad \boxed{\frac{1}{\phi} \:=\:\phi - 1}$

We have: .$\displaystyle \frac{1}{\phi} \;=\;\phi - 1$

Square: .$\displaystyle \frac{1}{\phi^2} \;=\;(\phi - 1)^2 \;=\;\phi^2 - 2\phi + 1 \;=\;(\phi+1) - 2\phi + 1 \;=\;2 - \phi$

Square: .$\displaystyle \frac{1}{\phi^4} \;=\;(2-\phi)^2 \;=\;4 - 4\phi + \phi^2 \;=\;4 - 4\phi + (\phi + 1) \;=\;5-3\phi$

Square: .$\displaystyle \frac{1}{\phi^8} \;=\;(5-3\phi)^2 \;=\;25 - 30\phi + 9\phi^2 \;=\;25 - 30\phi + 9(\phi+1) \;=\;34 - 21\phi$

Square: .$\displaystyle \frac{1}{\phi^{16}} \;=\;(34-21\phi)^2 \;=\;1156 - 1428\phi + 441\phi^2 \;=\; 1597 - 987\phi$

Hence: .$\displaystyle \frac{1}{\phi^{16}} \;=\;1597 - 987\left(\frac{1+\sqrt{5}}{2}\right) \;=\;\frac{2207 - 987\sqrt{5}}{2}$

Therefore: .$\displaystyle \sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}$
• Nov 24th 2012, 07:12 PM
Prove It
Re: Simplifying help
Quote:

Originally Posted by Soroban
Hello, gfbrd!

The original expression has a "minus": .$\displaystyle \sqrt[8]{\frac{2207 \:{\color{red}-}\: 987\sqrt{5}}{2}}$

You must mean: .$\displaystyle \frac{3\:{\color{red}-}\:\sqrt{5}}{2}$
. . which happens to equal: .$\displaystyle \frac{1}{\phi^2}$
. . where $\displaystyle \phi \,=\,\tfrac{1+\sqrt{5}}{2}$, the Golden Ratio.

It can be shown that: .$\displaystyle \frac{1}{\phi^{16}} \:=\:\frac{2207 - 987\sqrt{5}}{2}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I'd like to share a "trick" for raising $\displaystyle \phi$ to higher powers.

One of the root of the quadratic equation: .$\displaystyle x^2 - x - 1 \:=\:0$ .is $\displaystyle \phi.$

This means: .$\displaystyle \phi^2 - \phi - 1 \:=\:0 \quad\Rightarrow\quad \boxed{\phi^2 \:=\:\phi + 1}$

Divide by $\displaystyle \phi\!:\;\phi \:=\:1 + \frac{1}{\phi} \quad\Rightarrow\quad \boxed{\frac{1}{\phi} \:=\:\phi - 1}$

We have: .$\displaystyle \frac{1}{\phi} \;=\;\phi - 1$

Square: .$\displaystyle \frac{1}{\phi^2} \;=\;(\phi - 1)^2 \;=\;\phi^2 - 2\phi + 1 \;=\;(\phi+1) - 2\phi + 1 \;=\;2 - \phi$

Square: .$\displaystyle \frac{1}{\phi^4} \;=\;(2-\phi)^2 \;=\;4 - 4\phi + \phi^2 \;=\;4 - 4\phi + (\phi + 1) \;=\;5-3\phi$

Square: .$\displaystyle \frac{1}{\phi^8} \;=\;(5-3\phi)^2 \;=\;25 - 30\phi + 9\phi^2 \;=\;25 - 30\phi + 9(\phi+1) \;=\;34 - 21\phi$

Square: .$\displaystyle \frac{1}{\phi^{16}} \;=\;(34-21\phi)^2 \;=\;1156 - 1428\phi + 441\phi^2 \;=\; 1597 - 987\phi$

Hence: .$\displaystyle \frac{1}{\phi^{16}} \;=\;1597 - 987\left(\frac{1+\sqrt{5}}{2}\right) \;=\;\frac{2207 - 987\sqrt{5}}{2}$

Therefore: .$\displaystyle \sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}$

The OP already made it clear that the original post had a typo and should have been \displaystyle \displaystyle \begin{align*} \sqrt[8]{\frac{2207 \mathbf{+} 987\sqrt{5}}{2}} \end{align*} giving the correct result.
• Nov 25th 2012, 07:36 PM
gfbrd
Re: Simplifying help
Thanks for your help everyone, I'll see if I can figure something out from what soroban did