Oh! The roots!

Algebra Level 3

If one of the roots of the equation $x^{2}-px+q=0$ is m times the other root then $\frac{m}{1+m^{2}}$ is equal to

\frac{q}{q^{2}-2p}

\frac{p}{q^{2}-2p}

\frac{q}{p^{2}-2q}

\frac{p}{p^{2}-2q}

1 solution

Tom Engelsman
Sep 30, 2016

Let the quadratic equation in question have roots x = k and x = m*k, or:

(x - k)(x - mk) = x^2 - (m+1) kx + m k^2 = 0 (i).

Matching the coefficients up now gives:

(m+1) k = p (ii),
m
k^2 = q (iii).

Squaring (ii) produces m^2 + 2m + 1 = (p/k)^2 => m^2 + 1 = (p/k)^2 - 2m = (p/k)^2 - 2*(q/k^2) (iv).

Finally, the expression m/(m^2 + 1) becomes:

m/(m^2 + 1) = (q/k^2) / [(p^2 - 2q)/k^2] = q/(p^2 - 2q).