Question

If roots of the equation $a{x}^2+bx+c=0$ are $\alpha ,\beta$, then the equation whose roots are $\frac{1+\alpha }{1-\alpha },\frac{1+\beta }{1-\beta },$ where $\alpha \ne 1,\beta \ne 1$ is

• A. $a(x^2-1)+b(x-1{)}^2+c(x+1{)}^2=0$
• B. $a(x+1{)}^2+b(x^2-1)+c(x+1)=0$
• C. $a(x-1{)}^2+b(x^2-1)+c(x-1)=0$
• D. $a(x-1{)}^2+b(x^2-1)+c(x+1{)}^2=0$

Answer 1

The correct option is D $a(x-1{)}^2+b(x^2-1)+c(x+1{)}^2=0$

Let $\frac{1+\alpha }{1-\alpha }=y$
$\Rightarrow 1+\alpha =y-\alpha y\Rightarrow \alpha =\frac{y-1}{y+1}$
Now, $\alpha$ is the root of the equation $a{x}^2+bx+c=0$
$\Rightarrow a{\alpha }^2+b\alpha +c=0$
$\Rightarrow a{\left(\frac{y-1}{y+1}\right)}^2+b\left(\frac{y-1}{y+1}\right)+c=0$

Hence, the required equation is $a(x-1{)}^2+b(x^2-1)+c(x+1{)}^2=0$