Question

The ratio of the roots of the equation $a{x}^2+bx+c=0$ is same as the ratio of the roots of the equation $p{x}^2+qx+r=0$. If $D_1$ and $D_2$ are the discriminants of $a{x}^2+bx+c=0$ and $p{x}^2+qx+r=0$ respectively. Find the ratio of $D_1$ and $D_2$.

Answer 1

Given both equations have same ratio of roots

Roots of 1st equation :$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Ratio of roots of 1st equation :$\frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}$

Similarly for 2nd equation we can find ratio of roots. Both have same ratio of roots.

$\Rightarrow \frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=\frac{-q+\sqrt{q^2-4pr}}{-q-\sqrt{q^2-4pr}}$

Adding 1 on both sides :-

$\Rightarrow 1+\frac{-b+\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=1+\frac{-q+\sqrt{q^2-4pr}}{-q-\sqrt{q^2-4pr}}$

$\Rightarrow \frac{-2b}{-b-\sqrt{b^2-4ac}}=\frac{-2q}{-q-\sqrt{q^2-4pr}}$

$\Rightarrow \frac{-2b}{-b-\sqrt{D_1}}=\frac{-2q}{-q-\sqrt{D_2}}$

$\Rightarrow 2bq+2b\sqrt{D_1}=2qb+2q\sqrt{D_2}$

$\Rightarrow \frac{D_1}{D_2}=\frac{b^2}{q^2}$