How to solve for $ x$ when $x^2 \times  a = (x-10)^2 \times b$ 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.)
 A: 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. 
A: $\textbf{Hint}$ : 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 :
 $x = \frac{20b \pm \sqrt{(20b)^2 - 4(100)b(b - a)}}{2(b - a)}$
 $= \frac{20b \pm \sqrt{400ab}}{2(b-a)} = \frac{20b \pm 20\sqrt{ab}}{2(b-a)}$
 $= 10 \frac{b \pm \sqrt{ab}}{b - a}$

A: 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.
