<s>[INST] Help Solving Equation I've been trying to solve this equation for a few hours, but somehow I'm stuck somewhere and I could use some help.
$\frac{1} {x^2} = -\frac {2(x-a)} {a^3} + \frac {1} {a^2}$
I know from the book that there are two solutions (this is the equation of a tangent line to $x^2$). One solution is $x = a$:
$\frac{1} {x^2} - \frac {1} {a^2} = -\frac {2(x-a)} {a^3}$
The second solution is $x = - \frac {a} {2}$, but I don't know how to get to it.
 [/INST] As lcm$(x^2,a^3,a^2)=x^2a^3$ multiplying either sides by that and rearranging we get
$$a^3-ax^2+2x^2(x-a)=0$$
$$2x^2(x-a)-a(x^2-a^2)=0$$
$$(x-a)\{2x^2-a(x+a)\}=0$$
If $\displaystyle x-a\ne0,2x^2-ax-a^2=0$
Now, $\displaystyle2x^2-ax-a^2=2x^2-2ax+ax-a^2=2x(x-a)+a(x-a)=(x-a)(2x+a)$
But we have already assumed $x-a\ne0$
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