Question

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

• A. $\left(\frac{\alpha }{1-\alpha }\right),\left(\frac{\beta }{1-\beta }\right)$
• B. $\left(\frac{1-\alpha }{\alpha }\right),\left(\frac{1-\beta }{\beta }\right)$
• C. $\left(\frac{\alpha }{1+\alpha }\right),\left(\frac{\beta }{1+\beta }\right)$
• D. $\left(\frac{1+\alpha }{\alpha }\right),\left(\frac{1+\beta }{\beta }\right)$
  

Answer 1

The correct option is C $\left(\frac{\alpha }{1+\alpha }\right),\left(\frac{\beta }{1+\beta }\right)$

Since $\alpha$ and $\beta$ are the roots of ${ax}^2+bx+c=0,$

therefore

$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$

The equation ${ax}^2-bx(x-1)+c{(x-1)}^2=0$ can be written as $x^2(a-b+c)+(b-2c)x+c=0$

Sum of the roots of this equation is

$S=-\frac{b-2c}{a-b+c}=\frac{-b+2c}{a-b+x}=\frac{-\frac{b}{a}+\frac{2c}{a}}{1-\frac{b}{a}+\frac{c}{a}}$

$\Rightarrow S=\frac{\alpha +\beta +2\alpha \beta }{1+\alpha +\beta +\alpha \beta }=\frac{\alpha }{\alpha +1}+\frac{\beta }{\beta +1}$

Product of the roots $=\frac{c}{a-b+c}$

$\Rightarrow P=\frac{\frac{c}{a}}{1-\frac{b}{a}+\frac{c}{a}}$

$\Rightarrow P=\frac{\alpha \beta }{1+\alpha +\beta +\alpha \beta }=\frac{\alpha }{\alpha +1}.\frac{\beta }{\beta +1}$

Thus, ${ax}^2-bx(x-1)+c{(x-1)}^2=0$

has $\frac{\alpha }{\alpha +1},\frac{\beta }{\beta +1}$ as its two roots.