Question

If $p_1,p_2$ are the roots of the quadratic equation $a{x}^2+bx+c=0$ and $q_1,q_2$ are the roots of the quadratic equation $c{x}^2+bx+a=0(a,b,c\in R)$ such that $p_1,q_1,p_2,q_2$ are in A.P. of distinct terms, then $\frac{a}{c}$ equals

• A. -1
• B. 1
• C. $\frac{1}{2}$
• D. 2

Answer 1

Answer: (A)

Given that $p_1,q_1,p_2{q}_2$ are in A.P.
$\therefore {\left(p_2-p_1\right)}^2={\left(q_2-q_1\right)}^2$
$\Rightarrow {\left(p_2+p_1\right)}^2-4p_1{p}_2={\left(q_2+q_1\right)}^2-4q_1{q}_2$
$\Rightarrow {\left(\frac{-b}{a}\right)}^2-4\left(\frac{c}{a}\right)={\left(\frac{-b}{c}\right)}^2-4\left(\frac{a}{c}\right)$
$\Rightarrow \frac{b^2-4ac}{a^2}=\frac{b^2-4ac}{c^2}$
Since $b^2-4ac$ is the discriminant of both the equations and roots are different
$\therefore b^2\ne 4ac$
$\therefore a^2=c^2\Rightarrow a=c$ (Not possible because two quadratic equations become identical)
or $a=-c\Rightarrow \frac{a}{c}=-1$