Question

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

Answer 1

The roots of quadratic equation $a{x}^2+bx+c=0$ are $\alpha$ and $\beta$ such that $\frac{1}{\alpha }+\frac{1}{\beta },\alpha +\beta$ and ${\alpha }^2+{\beta }^2$ are in geometric progression.

We have $(\dot{\alpha }+\beta {)}^2=(\frac{1}{\alpha }+\frac{1}{\beta })({\alpha }^2+{\beta }^2)$

$\Rightarrow (\alpha +\beta {)}^2=(\frac{1}{\alpha }+\frac{1}{\beta })\left[\right(\alpha +\beta {)}^2-2\alpha \beta ]$

Substituting $\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$ we have

$\frac{b^2}{a^2}=\frac{-b}{c}(\frac{b^2}{a^2}-\frac{2c}{a})$

$\Rightarrow c{b}^2+b(b^2-2ac)=0$

$b\ne 0,\therefore bc+b^2-2ac=0$

$a,b,c$ are in AP, $\therefore b=\frac{a+c}{2}$

$\therefore$ we have $\frac{(a+c)c}{2}+{\left(\frac{a+c}{2}\right)}^2-2ac=0$

$\Rightarrow a^2-4ac+3c^2=0\Rightarrow (a-c)(a-3c)=0$

$a\ne c\therefore a=3c\therefore \frac{a}{c}=3$