Question

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

• A. $ac{x}^2-bx+1=0$
• B. $x^2-acx+bc+1=0$
• C. $ac{x}^2+bx-1=0$
• D. $x^2+acx-bc+11=0$
• E. $ac{x}^2-bx-11=0$

Answer 1

Answer: (A)

$\because$ $\alpha ,\beta$ are roots of equation, $a{x}^2+bx+c=0$ $(c\ne 0)$
$\therefore$ $\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}$
$\therefore$ Required equation is $x^2-\left(\frac{1}{a\alpha +b}+\frac{1}{\alpha \beta +b}\right)x+\left(\frac{1}{a\alpha +b}.\frac{1}{a\beta +b}\right)$
$\Rightarrow$ $x^2-\left(\frac{a(\alpha +\beta )+2b}{a^2\alpha \beta +ab(\alpha +\beta )+b^2}\right)x$ $+\left(\frac{1}{a^2\alpha \beta +ab(\alpha +\beta )+b^2}\right)=0$
$\Rightarrow$ $x^2-\left(\frac{a\left(-\frac{b}{a}\right)+2b}{a^2.\frac{c}{a}+ab\left(-\frac{b}{a}\right)+b^2}\right)x$ $+\frac{1}{a^2.\frac{c}{a}+ab\left(-\frac{b}{a}\right)+b^2}=0$
$\Rightarrow$ $x^2-\frac{b}{ac}x+\frac{1}{ac}=0$
$\Rightarrow$ $ac{x}^2-bx+1=0$