Question

If $\alpha$ and $\beta$ are the roots of $a{x}^2+bx+c=0$ then find the roots of the equation $a{x}^2-bx(x-1)+c{(x-1)}^2=0$

Answer 1

$\because a{x}^2-bx(x-1)+c{(x-1)}^2=0$ ........$\left(1\right)$

$\Rightarrow a{\left(\frac{x}{x-1}\right)}^2-b\left(\frac{x}{x-1}\right)+c=0$

$\Rightarrow a{\left(\frac{x}{1-x}\right)}^2+b\left(\frac{x}{1-x}\right)+c=0$

Now, $\alpha ,\beta$ are the roots of $a{x}^2+bx+c=0$ ........$\left(2\right)$

$\therefore a{\alpha }^2+b\alpha +c=0$ ........$\left(3\right)$

$a{\beta }^2+b\beta +c=0$ ........$\left(4\right)$

Comparing coefficient of $x$ from equation $\left(3\right)$ and $\left(4\right)$ with $\left(2\right)$

We get, $\alpha =\frac{x}{1-x}$ and $\beta =\frac{x}{1-x}$

$\Rightarrow x=\frac{\alpha }{1+\alpha }$ and $x=\frac{\beta }{1+\beta }$

Hence $\frac{\alpha }{1+\alpha }$ and $\frac{\beta }{1+\beta }$ are the roots of equation $\left(1\right)$