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

• 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

The correct option is A $\frac{a^2}{r^2}$

From the equation, $a{x}^2+bx+c=0$ if $x_1,x_2$ are its roots then

Sum of the roots $=x_1+x_2=-\frac{b}{a}$ and product of the roots $=x_1{x}_2=\frac{c}{a}$

From the equation, $p{x}^2+qx+r=0$ if $x_3,x_4$ are its roots then

Sum of the roots$=x_3+x_4=-\frac{q}{p}$ and product of the roots $=x_1{x}_2=\frac{r}{p}$

Given: $x_1,x_2,\frac{1}{x_3},\frac{1}{x_4}$ are in A.P

$\therefore x_2-x_1=\frac{1}{x_4}-\frac{1}{x_3}$

Squaring both sides, we get

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

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

Using the formula ${(a-b)}^2={(a+b)}^2-4ab$ we get

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

On substituting the above values in $\left(1\right)$ we get

$\frac{\frac{b^2}{a^2}-\frac{4c}{a}}{\frac{q^2}{p^2}-\frac{4r}{p}}=\frac{1}{\frac{r^2}{p^2}}$

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

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

Like terms in the L.H.S and R.H.S get cancelled and we get

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

Hence option $A$ is the answer.