Question

Suppose $x_1,x_2$ be the roots of $a{x}^2+bx+c=0$ and $x_3,x_4$ be the roots of $p{x}^2+qx+r=0$

If $x_1,x_2,\frac{1}{x_3},\frac{1}{x_4}$ are in A.P., then $\frac{b^2-4ac}{q^2-4pr}$ equals,

• A. $\frac{a^2}{r^2}$
• B. $\frac{b^2}{q^2}$
• C. $\frac{c^2}{p^2}$
• D. $\frac{a^2}{p^2}$

Answer 1

Answer: (A)

$\frac{a^2}{r^2}$

$x_1+x_2=\frac{-b}{a}and{x}_1.x_2=\frac{c}{a}$,

$x_3+x_4=\frac{-q}{p}and{x}_3.x_4=\frac{r}{p}$,

$x_2-x_1=\frac{1}{x_4}-\frac{1}{x_3}\Rightarrow (x_2-x_1{)}^2=\frac{(x_3-x_4{)}^2}{(x_4{x}_3{)}^2}$

$\Rightarrow \frac{(x_2-x_1{)}^2}{(x_4-x_3{)}^2}=\frac{1}{(x_3{x}_4{)}^2}$

$\Rightarrow \frac{(x_2+x_1{)}^2-4x_1{x}_2}{(x_3+x_4{)}^2-4x_3{x}_4}=\frac{1}{(x_3{x}_4{)}^2}$

On putting values we get

$\frac{b^2-4ac}{q^2-4pr}=\frac{a^2}{r^2}$

Hence option ${}^{\prime }{A}^{\prime }$ is the answer.