Discussion Forum

1)

If $\alpha$ and $\beta$ are the roots of the equation  $ax^{2}+bx+c=0$ and the equation having roots $\frac{1-\alpha}{\alpha}$

and $\frac{1-\beta}{\beta}$ is $px^{2}+qx+r=0$ then r=

Answer:

Option C

Explanation:

We have,

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

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

  Equations whose roots are $\frac{1-\alpha}{\alpha}$ and $\frac{1-\beta}{\beta}$ are

  $x^{2}-\left(\frac{1-\alpha}{\alpha}+\frac{1-\beta}{\beta}\right)x+\left(\frac{1-\alpha}{\alpha}\right)\left(\frac{1-\beta}{\beta}\right)=0$

$\Rightarrow$    $x^{2}-\left(\frac{1}{\alpha}+\frac{1}{\beta}-2\right)x+\left(\frac{1-(\alpha+\beta)+\alpha\beta}{\alpha\beta}\right)=0$

$\Rightarrow$    $x^{2}-\left(\frac{\alpha+\beta-2\alpha\beta}{\alpha\beta}\right)x+\left(\frac{1-(\alpha+\beta)+\alpha\beta}{\alpha\beta}\right)=0$

$\Rightarrow$     $\alpha\beta x^{2}-(\alpha+\beta-2\alpha\beta)x+(1-(\alpha+\beta)+\alpha\beta)=0$

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

$\Rightarrow$     $cx^{2}+(b+2c)x+(a+b+c)=0$

Comparing with $px^{2}+qx+r=0$ we get

                                           r= a+b+c