How to extract values from an equation

**prettiestfriend** · June 9th 2009, 02:54 PM

How do I get n from this:

I get stuck at

v/R - 1/4 = 1/n^2

**HallsofIvy** · June 9th 2009, 03:30 PM

Originally Posted by prettiestfriend

How do I get n from this:

I get stuck at

v/R - 1/4 = 1/n^2

Go ahead and do the subtraction on the left. The "common denominator" is 4R so you have 4v/(4R)- R/(4R)= (4v- R)/(4R)= 1/n^2. Now take the reciprocal of both sides: 4R/(4v- R)= n^2. Finally, just take the square root of both sides.

**Deadstar** · June 9th 2009, 03:32 PM

$v = R \bigg{(}\frac{1}{4} - \frac{1}{n^2} \bigg{)}$

$=> \frac{v}{R} = \frac{1}{4} - \frac{1}{n^2}$

$=> \frac{v}{R} - \frac{1}{4} = -\frac{1}{n^2}$

$=> \frac{1}{4} - \frac{v}{R} = \frac{1}{n^2}$

Now $\frac{1}{4} - \frac{v}{R}= \frac{R - 4v}{4R}$ and then multiply both sides by $n^2$ ...

$=> n^2 \cdot \frac{R - 4v}{4R} = 1$

Then divide both sides by $\frac{R - 4v}{4R}$ ...

$n^2 = \frac{1}{\frac{R - 4v}{4R}} = \frac{4R}{R - 4v}$

Then take the square root of each side to get...

$n = \sqrt{\frac{4R}{R - 4v}}$

**Soroban** · June 9th 2009, 03:49 PM

Hello, prettiestfriend!

Solve for $n\!:\;\;v \;=\;R\left(\frac{1}{4} - \frac{1}{n^2}\right)$

Get rid of the fractions as soon as possible . . .

We have: . $v \;=\;\frac{R}{4} - \frac{R}{n^2}$

Multiply by $4n^2\!:\quad 4n^2(v) \;=\;{\color{red}\rlap{/}}4n^2\left(\frac{R}{{\color{red}\rlap{/}}4}\right) - 4{\color{red}\rlap{//}}n^2\left(\frac{R}{{\color{red}\rlap{//}}n^2}\right)$

. . and we have: . $4vn^2 \;=\;Rn^2 - 4R\quad\Rightarrow\quad Rn^2 - 4vn^2 \:=\:4R$

Factor: . $(R - 4v)n^2 \;=\;4R \quad\Rightarrow\quad n^2 \;=\;\frac{4R}{R - 4v}$

Take the square root: . $n \;=\;\pm\sqrt{\frac{4R}{R - 4v}}$

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