Question 70858
You're on the right track, we're going to complete the square
{{{x^2+m^2-2mx=cross(2mx-2mx)+(nx)^2}}}subtract 2mx from both sides (edit: sorry you're not supposed to subtract m^2 on the left hand side)
Now we need to find 2 terms which add to -2mx and multiply to m^2, these two are -m and -m, so the factors are
{{{(x-m)(x-m)=n^2*x^2}}}
{{{(x-m)^2=n^2*x^2}}}Which can be also written as this
{{{sqrt((x-m)^2)=sqrt(n^2*x^2)}}}Take square root of both sides
{{{x-m=nx}}}Subtract nx and add m to both sides
{{{x(1-n)=m}}}Factor out an x
{{{(x*cross(1-n))/cross(1-n)=m/(1-n)}}}Divide both sides by (1-n)
{{{x=m/(1-n)}}}
To verify this let m and n be any numbers (preferably small numbers to make it easy) and this should be equivalent to the given problem.
Check let m=1, n=2, and plug in x=1
{{{x=1/(1-2)=-1}}}
{{{(-1)^2+1^2=2*1*(-1)+(2*1)^2}}}
{{{2=2}}} So the answer checks out