Simplifying help

**gfbrd** · November 24th 2012, 09:07 AM

Hey there I need some help simplifying this problem
hope you guys can see it

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

**skeeter** · November 24th 2012, 09:17 AM

$\sqrt[8]{\frac{2207-987\sqrt{5}}{2}}$

what makes you think the expression can be simplified further than the form you've written?

**gfbrd** · November 24th 2012, 09:19 AM

because i know the simplified form

i just dont know how to go about simplifying it thats all

**skeeter** · November 24th 2012, 09:30 AM

care to share that "simplified form" ... because my CAS can't do it.

**gfbrd** · November 24th 2012, 09:36 AM

it should be
(3 + sqrt(5))/ 2

**skeeter** · November 24th 2012, 09:45 AM

uhh ...

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

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

???

**gfbrd** · November 24th 2012, 09:56 AM

oh sorry i made a mistake in typing the problem
it should be + 987sqrt(5) not subtract

**skeeter** · November 24th 2012, 10:14 AM

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.

**gfbrd** · November 24th 2012, 10:25 AM

oh alright thanks for helping

**Prove It** · November 24th 2012, 05:10 PM

Originally Posted by gfbrd

Hey there I need some help simplifying this problem
hope you guys can see it

Click image for larger version.

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ID: 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.

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

I'll have to think about how the top simplifies to $\displaystyle \begin{align*} 3 + \sqrt{5} \end{align*}$ ...

**Soroban** · November 24th 2012, 07:02 PM

Hello, gfbrd!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

**Prove It** · November 24th 2012, 07:12 PM

Originally Posted by Soroban

Hello, gfbrd!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The OP already made it clear that the original post had a typo and should have been $\displaystyle \begin{align*} \sqrt[8]{\frac{2207 \mathbf{+} 987\sqrt{5}}{2}} \end{align*}$ giving the correct result.

**gfbrd** · November 25th 2012, 07:36 PM

Thanks for your help everyone, I'll see if I can figure something out from what soroban did

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