Question

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

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

Answer 1

The correct option is D $\frac{\alpha }{\alpha -1},\frac{\beta }{\beta -1}$

There is a small error in the equation which should be $a{x}^2+bx+c=0$ , instead of $a{x}^2+bx+x=0$.

Now we know,

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

$\Rightarrow b=-(\alpha +\beta )a$ and $c=\left(\alpha \beta \right)a$

Let roots of equation $(a+b+c)x^2-(b+2c)x+c=0$ be ${\alpha }_1$ and ${\beta }_1$.

So,

${\alpha }_1+{\beta }_1=\frac{b+2c}{a+b+c}$ and ${\alpha }_1{\beta }_1=\frac{c}{a+b+c}$

$\Rightarrow {\alpha }_1+{\beta }_1=\frac{-(\alpha +\beta )a+2\left(\alpha \beta \right)a}{a-(\alpha +\beta )a+\left(\alpha \beta \right)a}$ and ${\alpha }_1{\beta }_1=\frac{\left(\alpha \beta \right)a}{a-(\alpha +\beta )a+\left(\alpha \beta \right)a}$

$\Rightarrow {\alpha }_1+{\beta }_1=\frac{2\left(\alpha \beta \right)-(\alpha +\beta )}{1-(\alpha +\beta )+\left(\alpha \beta \right)}$ and ${\alpha }_1{\beta }_1=\frac{\alpha \beta }{1-(\alpha +\beta )+\left(\alpha \beta \right)}$

Let,

${\alpha }_1+{\beta }_1=\frac{-B}{A}$ and ${\alpha }_1{\beta }_1=\frac{C}{A}$

So,roots ${\alpha }_1$ or ${\beta }_1$=$\frac{-B\pm \sqrt{B^2-4AC}}{2A}$

Now, let

$\sqrt{D}=\sqrt{B^2-4AC}=\sqrt{{\{2\left(\alpha \beta \right)-(\alpha +\beta )\}}^2-4\left(\alpha \beta \right)\{1-(\alpha +\beta )+\left(\alpha \beta \right)\}}$

$\Rightarrow \sqrt{D}=\alpha -\beta$

So, ${\alpha }_1$ or ${\beta }_1=\frac{-B\pm \sqrt{D}}{2A}$

So, ${\alpha }_1=\frac{2\alpha \beta -(\alpha +\beta )+\alpha -\beta }{1-(\alpha +\beta )+\alpha \beta }$

$\Rightarrow {\alpha }_1=\frac{2\left(\alpha \beta \right)-2\beta }{2\{1-(\alpha +\beta )+\left(\alpha \beta \right)\}}$

$\Rightarrow {\alpha }_1=\frac{\beta }{\beta -1}$

Similarly,

${\beta }_1=\frac{2\left(\alpha \beta \right)-(\alpha +\beta )-\alpha +\beta }{2\{1-(\alpha +\beta )+\left(\alpha \beta \right)\}}$

$\Rightarrow {\beta }_1=\frac{\alpha }{\alpha -1}$