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 $a{x}^2+bx+c=0$ 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

Answer: 0003

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