Question

If $\alpha \&\beta$ are the roots of the equation $a{x}^2+bx+c=0,$ The equation whose roots are $1+\frac{1}{\alpha },1+\frac{1}{\beta }$ will be:

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

Answer 1

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

Given: $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$
To find: Equation with roots $1+\frac{1}{\alpha },1+\frac{1}{\beta }$

Let $y=1+\frac{1}{x}$
$\Rightarrow x=\frac{1}{y-1}$

Substituting $x=\frac{1}{y-1}\text{in}a{x}^2+bx+c=0,$ we get:
$a{\left(\frac{1}{y-1}\right)}^2+b\left(\frac{1}{y-1}\right)+c=0$

$\Rightarrow a+b(y-1)+c(y-1{)}^2=0$

$\Rightarrow a+by-b+c{y}^2-2cy+c=0$

$\Rightarrow c{y}^2+(b-2c)y+(a-b+c)=0$

Now, In terms of $x,$ we get the equation as:
$c{x}^2+(b-2c)x+(a-b+c)=0$ which is required equation.