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$1/[a\alpha +b]$and $1/[a\beta +b],$ is

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

Answer 1

The correct option is A

$ac{x}^2–bx+1=0$

Explanation for the correct option:

Find the equation :

Given $\alpha$and $\beta$are the roots of the equation $a{x}^2+bx+c=0(c\ne 0)$,

$\Rightarrow ɑ+\beta =–b/a,ɑ\beta =c/a$

The required equation is

$$\begin{gathered} x^2–\left[\frac{1}{(a\alpha +b)}+\frac{1}{(a\beta +b)}\right]x+\left[\frac{1}{(a\alpha +b)}\times \frac{1}{(a\beta +b)}\right]=0 \\ \Rightarrow x^2–\left[\frac{(a(ɑ+\beta )+2b)}{(a^2ɑ\beta +ab(ɑ+\beta )+b^2)}\right]x+\left[\frac{1}{(a^2ɑ\beta +ab(ɑ+\beta )+b^2)}\right]=0 \\ \Rightarrow x^2–\left[\frac{a\left({\displaystyle \frac{-b}{a}}\right)+2b)}{(a^2\left(\frac{c}{a}\right)+ab\left(\frac{-b}{a}\right)+b^2)}\right]x+\left[\frac{1}{(a^2\left(\frac{c}{a}\right)+ab\left(\frac{-b}{a}\right)+b^2)}\right]=0 \\ \Rightarrow x^2–\left(\frac{b}{ac}\right)x+\left(\frac{1}{ac}\right)=0 \\ \Rightarrow ac{x}^2–bx+1=0 \end{gathered}$$

Hence the correct option is A.