Question

Let $a , b$ and $c$ are all different and non-zero real numbers (taken in that order) are in arithmetic progression. If the roots of quadratic equation $ax ^2+ bx + c =0$ are $\alpha$ and $\beta$ such that $\alpha^{-1}+\beta^{-1}, \alpha+\beta, \alpha^2+\beta^2$ (taken in that order) are in geometric progression, then find the value of $\left(\frac{ a }{ c }\right)$.

Answer 1

We have
$(\alpha+\beta)^2=\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\left(\alpha^2+\beta^2\right) \Rightarrow \frac{b^2}{a^2}=\left(\frac{-b}{c}\right)\left(\frac{b^2}{a^2}-\frac{2 c}{a}\right) $
$\Rightarrow cb ^2=- b ^3+2 abc \Rightarrow cb ^2+ b \left( b ^2-2 ac \right)=0$
$\text { As } b \neq 0 \text {, so } $
$b c+b^2-2 a c=0 $....(i)
Also, $ b =\frac{ a + c }{2}$
$\therefore$ Using (ii) in (i), we get
$c \left(\frac{ a + c }{2}\right)+\left(\frac{ a + c }{2}\right)^2-2 ac =0 \Rightarrow a ^2-4 ac +3 c ^2=0 \Rightarrow( a - c )( a -3 c )=0$
As $a \neq c$
$\Rightarrow \frac{ a }{ c }=3 $