Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0,$ then the roots of the equation $c{x}^2+(b^2-2c)x+c=0$ are

Answer 1

$\alpha$ & $\beta$ are roots of $x^2+bx+c=0$
$\Rightarrow \alpha +\beta =-b$
$\alpha \beta =c$
Let us now consider $c{x}^2+(b^2-2c)x+c=0$
Sum of roots =$-\frac{(b^2-2c)}{c}=\frac{2c-b^2}{c}=2-\frac{b^2}{c}$
$=2-\frac{(\alpha +\beta {)}^2}{\alpha \beta }$
$=2-\frac{{\alpha }^2+2\alpha \beta +{\beta }^2}{\alpha \beta }$
$=\frac{2\alpha \beta -{\alpha }^2-2\alpha \beta -{\beta }^2}{\alpha \beta }$
$=\frac{-{\alpha }^2-{\beta }^2}{\alpha \beta }$
$=-\frac{\alpha }{\beta }-\frac{\beta }{\alpha }$
$=(-\frac{\alpha }{\beta })+(-\frac{\beta }{\alpha })$
Product of roots is $\frac{c}{c}=1$
Since $-\frac{\alpha }{\beta }\times -\frac{\beta }{\alpha }=1$
Thus, the roots are $(-\frac{\alpha }{\beta })$ & $(-\frac{\beta }{\alpha })$ as they satisfy the above condition of product of roots.
Hence, $-\frac{\alpha }{\beta }$ and $-\frac{\beta }{\alpha }$ are the roots of the equation $c{x}^2+(b^2-2c)x+c=0$.