If k be the ratio of the roots of the equation $ \displaystyle x^{2}-px+q=0 $ , the value of $ \displaystyle \frac{k}{1+k^{2}} $ is
- $ \displaystyle \frac{q^{2}-2p}{p} $
- $ \displaystyle \frac{q}{p^{2}-2q} $
- $ \displaystyle \frac{p}{q^{2}-2p} $
- $ \displaystyle \frac{p}{p^{2}-2q} $
Let roots be a and ka. Sum = a(1+k) = p. Product = ka^2 = q. So a = p/(1+k). Substituting into product: k(p/(1+k))^2 = q. k*p^2 = q(1+k)^2 = q(1 + 2k + k^2). Rearranging: k*p^2 = q + 2kq + kq^2. k(p^2 - 2q - q^2) = q. This leads to k/(1+k^2) = q/(p^2-2q).
Let the roots be k times alpha and alpha, where their sum gives p equals alpha times (k plus 1) and their product gives q equals k times alpha squared. We square the sum equation to get p squared equals alpha squared times (k squared plus 2k plus 1), and then substitute alpha squared as q divided by k. Multiplying out the terms results in k times p squared equals q times k squared plus 2q times k plus q, which rearranges to q times k squared plus (2q minus p squared) times k plus q equals 0. Dividing the entire equation by k squared yields q plus (2q minus p squared) divided by k plus q divided by k squared equals 0, and dividing by q gives k divided by (1 plus k squared) equals (p squared minus 2q) divided by q, which simplifies to k divided by (1 plus k squared) equals q divided by (p squared - 2q).