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I'm trying to figure this one out, and I'm stuck. I'm going:
$x^2 \times a = (x-10)^2 \times b $
$x^2 \times a/b = (x-10)^2 /b $
$a/b = \frac{(x-10)^2} { x^2} / x^2$
$\sqrt{a/b} = (x-10)/x$

I'm not sure how I proceed or if this is even the best way to start. (I do fine with everything up to linear algebra.)

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If you want to proceed, let $c = \sqrt{a/b}$, then $(x-10)/x = c$ $\implies cx = x - 10$ $\implies cx - x = -10$ $\implies x(c-1) = -10$ $\implies x = -10/(\sqrt{a/b} - 1)$. (edit: I wrote "if" in bold because of the $\pm$ you have to add. See Andre's answer.) –  user2468 Sep 5 '12 at 20:39
    
@draks: OP's approach avoids the quadratic. –  Ross Millikan Sep 5 '12 at 20:41
    
@RossMillikan At the expense of forgetting to add a $\pm$ to his formula, of course. –  Thomas Andrews Sep 5 '12 at 20:43
    
Awesome thanks. I did forget that when you take the square root it becomes +/-. It's been decades since I did this stuff in an academic setting (where the process mattered as much as the answer) so thanks for the reminder. Also thanks for the idea of substituting everything with a new variable to keep it simpler. (I'd also forgotten the trick that (a + b) x = ax + bx. Ironic since I teach people about FOIL.) –  Chuck P Sep 6 '12 at 1:57
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3 Answers

up vote 2 down vote accepted

I will assume that $a$ and $b$ are $\ge 0$, but are not both $0$. Take the square root(s). We get $$\sqrt{a} \,x=\pm\sqrt{b}(x-10).$$ Now we have two linear equations. Deal with them separately.

The equation $\sqrt{a}x=\sqrt{b}(x-10)$ can be rewritten as $\sqrt{a}\,x =\sqrt{b}\, x-10\sqrt{b}$ and then as $10\sqrt{b}=(\sqrt{b}-\sqrt{a})x$. If $a\ne b$, it has solution $$x=\frac{10\sqrt{b}}{\sqrt{b}-\sqrt{a}} .$$ If $a=b$ the linear equation has no solution. The other equation is dealt with similarly. It gives after a while $$x=\frac{10\sqrt{b}}{\sqrt{a}+\sqrt{b}} .$$

Remark: As written, the equation was awfully close to a linear (that is, nice) equation. If I am that close, I prefer to go directly for the prize.

For completeness, we deal with the other possibilities for $a$ and $b$. If $a$ and $b$ are both $\le 0$, but are not both $0$, use the equivalent equation $|a|x^2=|b|(x-10)^2$. If $a$ is positive and $b$ is negative, or the other way around, there is no solution. And if $a$ and $b$ are both $0$, then every real number $x$ is a solution.

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Thanks, this is what I was looking for! The original problem I had decided to solve was "two objects leave the same point 10 seconds apart, accelerating at different rates, when does object B catch up?" In that case b>a and both are positive. I could graph it and find an intersection to within a value of 1 (or just program a loop to find the right value of x by brute force) but I wanted to do it "right" this time. –  Chuck P Sep 6 '12 at 1:06
    
@ChuckP: In that case, since $x\gt 10$, we have $\sqrt{a}\,x=\sqrt{b}\,(x-10)$, and the first solution, the one with $\sqrt{b}-\sqrt{a}$ in the denominator, is the physically correct one. –  André Nicolas Sep 6 '12 at 1:13
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$\textbf{Hint}$ : Try to put it into the form $Dx^2 + Ex + F$ and use the quadratic formula to find the solutions. Move cursor over the box for more details.

$ax^2 = b(x - 10)^2$
$ax^2 = b(x^2 - 20x + 100)$
$ax^2 = bx^2 - 20bx + 100b$
$0 = (b - a)x^2 - 20bx + 100b$
Now using the quadratic formula :
$x = \frac{20b \pm \sqrt{(20b)^2 - 4(100)b(b - a)}}{2(b - a)}$
$= \frac{20b \pm \sqrt{400ab}}{2(b-a)} = \frac{20b \pm 20\sqrt{ab}}{2(b-a)}$
$= 10 \frac{b \pm \sqrt{ab}}{b - a}$

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Hint: now consider $\sqrt {\frac ab}=c$ and you have $c=\frac {x-10}x$, a linear equation. So multiply by $x$, collect the terms that include $x$, etc.

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No, you have $\frac{x-10}{x} = \pm c$ –  Thomas Andrews Sep 5 '12 at 20:43
    
@ThomasAndrews: True. I saw the squares and realized $\frac ab$ had to be positive, but we could have $c \lt 0$ if $0 \lt x \lt 10$ –  Ross Millikan Sep 5 '12 at 21:24
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