Simplifying help

Printable View

November 24th 2012, 09:07 AM
gfbrd

1 Attachment(s)

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
November 24th 2012, 09:17 AM
skeeter

Re: Simplifying help

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

what makes you think the expression can be simplified further than the form you've written?
November 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
November 24th 2012, 09:30 AM
skeeter

Re: Simplifying help

care to share that "simplified form" ... because my CAS can't do it.
November 24th 2012, 09:36 AM
gfbrd

Re: Simplifying help

it should be
(3 + sqrt(5))/ 2
November 24th 2012, 09:45 AM
skeeter

Re: Simplifying help

uhh ...

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

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

???
November 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
November 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.
November 24th 2012, 10:25 AM
gfbrd

Re: Simplifying help

oh alright thanks for helping
November 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.

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

I'll have to think about how the top simplifies to $\displaystyle \begin{align*} 3 + \sqrt{5} \end{align*}$ ...
November 24th 2012, 07:02 PM
Soroban

Re: Simplifying help

Hello, gfbrd!

Quote:

$\text{It should be }\:\frac{3 + \sqrt{5}}{2}$ . . . . 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: . $\sqrt[8]{\frac{2207 - 987\sqrt{5}}{2}} \;=\;\frac{1}{\phi^2} \;=\;\frac{3 - \sqrt{5}}{2}$
November 24th 2012, 07:12 PM
Prove It

Re: Simplifying help

Quote:

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.
November 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