Question

If $p _1, p _2$ are the roots of the quadratic equation $a x^2+b x+c=0$ and $q _1, q _2$ are the roots of the quadratic equation $cx ^2+ bx + a =0( a , b , c \in R )$ 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)^2-4 p _1 p _2=\left( q _2+ q _1\right)^2-4 q _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) $
$\Rightarrow \frac{ b ^2-4 ac }{ a ^2}=\frac{ b ^2-4 ac }{ c ^2}$
Since $b^2-4 a c$ is the discriminant of both the equations and roots are different
$\therefore b^2 \neq 4 a c$
$\therefore a^2=c^2 \Rightarrow a=c$ (Not possible because two quadratic equations become identical)
or $ a=-c \Rightarrow \frac{a}{c}=-1$