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, then $D_1:D_2$ is equal to

• A. $\frac{a^2}{p^2}$
• B. $\frac{b^2}{q^2}$
• C. $\frac{c^2}{r^2}$
• D. None of these

Answer 1

Answer: (B)

$\frac{b^2}{q^2}$

Let ${\alpha }_1,{\beta }_1$ be the roots of $a{x}^2+bx+c=0$ and ${\alpha }_2,{\beta }_2$ be the roots of $p{x}^2+qx+r=0$. Then

$\frac{{\alpha }_1}{{\beta }_1}=\frac{{\alpha }_2}{{\beta }_2}$ (given)

$\frac{{\alpha }_1+{\beta }_1}{{\alpha }_1-{\beta }_1}=\frac{{\alpha }_2+{\beta }_2}{{\alpha }_2-{\beta }_2}$ (Applying componendo and dividendo)

$\frac{{({\alpha }_1+{\beta }_1)}^2}{{({\alpha }_1-{\beta }_1)}^2}=\frac{{({\alpha }_2+{\beta }_2)}^2}{{({\alpha }_2-{\beta }_2)}^2}$

$\frac{{({\alpha }_1+{\beta }_1)}^2}{{({\alpha }_1+{\beta }_1)}^2-4{\alpha }_1{\beta }_1}=\frac{{({\alpha }_2+{\beta }_2)}^2}{{({\alpha }_2+{\beta }_2)}^2-4{\alpha }_2{\beta }_2}$

$\frac{\frac{b^2}{a^2}}{\frac{b^2-4ac}{a^2}}=\frac{\frac{q^2}{p^2}}{\frac{q^2-4rp}{p^2}}$

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

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