A Polynomials Question..?

Question

The roots of an equation are the reciprocals of the roots of the equation:
x^2+2ax-c^2=0
Find that equation..
Thanks..

The answer given is (c^2x^2)-2ax-1=0 but I have no idea how to get it..

Answer 1

let f(x)=x^2+2ax-c^2. By the quadratic equation, the roots of f are
x=(-2a±√(4a^2-4c^2))/2= -a±√(a^2-c^2)

We want to find the equation for g(x), where the roots of g are the recipricals of the roots of f. Thus the roots of g are 1/(-a±√a^2+c^2). and g(x)=m • (x-1/(-a+√a^2+c^2)) • (x-1/(-a-√a^2+c^2)), were m is a constant.

Expanding g, we get:
g(x)=m • {x^2-(1/(-a+√a^2+c^2)+ 1/(-a-√a^2+c^2)) x + [1/(-a±√a^2+c^2)] • [1/(-a-√a^2+c^2)]} = m • {x^2 - [(-a+√a^2+c^2) + (-a-√a^2+c^2)/ (a^2-(a^2+c^2)]x +1/(a^2-(a^2+c^2)} = m • {x^2 - [(-2a/(-c^2)]x +1/(-c^2)}.

Since we want integer coefficients, let m=c^2

Thus g(x)=c^2 • (x^2 -(2a/c^2)x -1/c^2)= c^2x^2-2ax-1.

To Nu$clip: The negatives come from the -c^2 in the first equation.


There of course is a much easier way of doing this:
It is obvious that x=0 is not a solution of this eqation, thus we can divide by x.
Take the original equation and devide by x^2

This gives 1+2a(1/x)- c^2(1/x^2) =0(1/x^2)=0. Therefore if x is a solution of the first equation, x is a solution of the second equation. Let z=1/x, then the second equation is 1+2az-c^2z^2=0 and 1/x is a solution to this equation. Multiply by -1 so that the leading coefficient of z^2 is positve, and we have c^2z^2-2az-1=0. This equation has roots that are recipricals of the first equation. Of course you can replace z with x, but it doesn't matter.