Question

If $\alpha$ and $\beta$ are roots of the equation $a{x}^2+bx+c=0$, then the roots of the equation $a(2x+1{)}^2-b(2x+1\left)\right(3-x)+c(3-x{)}^2=0$ are

• A. $\frac{2\alpha +1}{\alpha -3},\frac{2\beta +1}{\beta -3}$
• B. $\frac{3\alpha +1}{\alpha -2},\frac{3\beta +1}{\beta -2}$
• C. $\frac{2\alpha -1}{\alpha -2},\frac{2\beta +1}{\beta -2}$
• D. none of these

Answer 1

The correct option is B $\frac{3\alpha +1}{\alpha -2},\frac{3\beta +1}{\beta -2}$

$a(2x+1{)}^2-b(2x+1\left)\right(3-x)+c(3-x{)}^2=0$

$a{\left(\frac{2x+1}{3-x}\right)}^2-b\frac{2x+1}{3-x}+c=0$
If, $\frac{2x+1}{x-3}=y$,

Then the equation can be expressed as :

$a{y}^2+by+c=0$

This equation will have roots $\alpha ,\beta$.

$y=\alpha \Rightarrow \frac{2x+1}{x-3}=\alpha$

$\Rightarrow 2x+1=\alpha x-3\alpha$

$\Rightarrow x(2-\alpha )=-1-3\alpha$

$\Rightarrow x=\frac{1+3\alpha }{\alpha -2}$

Similarly , $y=\beta \Rightarrow x=\frac{1+3\beta }{\beta -2}$

Hence, the roots of the equation are $\frac{1+3\alpha }{\alpha -2},\frac{1+3\beta }{\beta -2}$