Question

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0,a\ne 0$. If $1,\alpha +\beta ,\alpha \beta$ are in A.P. and $\frac{1}{\alpha },\frac{1}{2},\frac{1}{\beta }$ are also in A.P., then the value of $\frac{{\alpha }^2+{\beta }^2-2{\alpha }^2{\beta }^2}{{\alpha }^2+{\beta }^2}$ is

Answer 1

$\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$
Then using sum and product of roots, we get
$\alpha +\beta =-\frac{b}{a}\alpha \beta =\frac{c}{a}$
Now,
$1,\alpha +\beta ,\alpha \beta$ are in A.P., so
$1+\alpha \beta =2(\alpha +\beta )\Rightarrow a+2b+c=0\cdots \left(1\right)$
Also,
$\frac{1}{\alpha },\frac{1}{2},\frac{1}{\beta }$ are in A.P., so
$\frac{1}{\alpha }+\frac{1}{\beta }=1\Rightarrow \frac{\alpha +\beta }{\alpha \beta }=1\Rightarrow -b=c\cdots \left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$, we get
$a=-b=c$
Then the equation becomes,
$a(x^2-x+1)=0\Rightarrow x^2-x+1=0(\because a\ne 0)$
Therefore,
$\frac{{\alpha }^2+{\beta }^2-2{\alpha }^2{\beta }^2}{{\alpha }^2+{\beta }^2}=1-\frac{2}{{\alpha }^2+{\beta }^2}=1-\frac{2}{(\alpha +\beta {)}^2-2\alpha \beta }=3$