If the roots of the equation $\displaystyle a_{1}x^{2}+a_{2}x+a_{3}=0$ are in the ratio $\displaystyle r_{1}:r_{2}$ then
- $\displaystyle r_{1}.r_{2}.a_{1}^{2}=(r_{1}+r_{2})a_{3}^{2}$
- $\displaystyle r_{1}.r_{2}.a_{2}^{2}=(r_{1}+r_{2})^{2}a_{1}.a_{3}$
- $\displaystyle r_{1}.r_{2}.a_{2}=(r_{1}+a_{2})^{2}a_{1}.a_{3}$
- $\displaystyle r_{1}.r_{2}.a_{1}^{2}=(r_{1}+r_{2})^{2}a_{3}.a_{2}$
For roots alpha and beta with ratio r1:r2, alpha/beta = r1/r2. Sum = -a2/a1, Product = a3/a1. Using (alpha+beta)^2 / (alpha*beta) = (r1+r2)^2 / (r1*r2), we get (a2^2/a1^2) / (a3/a1) = (r1+r2)^2 / (r1*r2), which simplifies to r1*r2*a2^2 = (r1+r2)^2 * a1*a3.
Let the roots be r1 times k and r2 times k; by Vieta's formulas, the sum of the roots is r1 plus r2 times k equals negative a2 divided by a1, and the product of the roots is r1 times r2 times k squared equals a3 divided by a1. Squaring the sum equation gives r1 plus r2 squared times k squared equals a2 squared divided by a1 squared. By substituting k squared from the product equation into the squared sum equation, we get r1 plus r2 squared times a3 divided by r1 times r2 times a1 equals a2 squared divided by a1 squared, which rearranges to r1 times r2 times a2 squared equals r1 plus r2 squared times a1 times a3.