How to solve this proportion

Question

Okay I think I'm just having a major brain block, but I need help solving this proportion for my physics class.

$$\frac {6.0\times 10^{-6}}{ x^2} = \frac {2.0\times 10^{-6}}{ (x-20)^2}$$

What's confusing me is the solution manual to this problem lists writing the proportion as,

$$\frac {(x-20)^2} { x^2} = \frac {2.0\times 10^{-6}}{ 6.0\times 10^{-6}}$$

and then proceeds to solve the problem from there... but that doesn't seem right to me. Usually you would cross multiply a proportion and solve, but they seemed to do some illegal math or something. Could you guys work me through how to solve this? This answer is 47 by the way.

Answer 1

It's only $47$ in physics. In mathematics, it would be

$$ \frac{20\sqrt{3}}{\sqrt{3}-1} = 47.320\ldots $$

:-)

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Anyway: You start off with a proportion that can be written, generally, as

$$ \frac{a}{b} = \frac{c}{d} $$

The book then proceeds to rewrite this as

$$ \frac{d}{b} = \frac{c}{a} $$

That the two are equivalent (provided $a \not= 0$) can be seen by multiplying both sides of the first equation by $d$, and then dividing by $a$. You can also see that both equations yield the same result after cross-multiplication.

Answer 2

The proportion they gave is correct. Simply divide both sides of the equation you started with by $6\cdot10^{-6}$ and multiply both sides by $(x-20)^2$.

So we have $$ \frac{(x-20)^2}{x^2} = \frac{2.0\cdot10^{-6}}{6.0\cdot10^{-6}}=\frac{1}{3}.$$ The left-hand side becomes $$\frac{(x-20)^2}{x^2}=\frac{x^2-40x+400}{x^2}=\frac{1}{3}.$$ Multiplying both sides by $x^2$, we get $$x^2-40x+400=\frac{x^2}{3} \quad\Rightarrow\quad\frac{2}{3}x^2-40x+400=0.$$ From here you use the quadratic equation to give you $x=10(3±\sqrt{3})$. It's physics, so you probably want the positive one, so $x=10(3+\sqrt{3})\approx47$.