Q52 / 160 · AP EAMCET 2023 (16 May Shift 1) · Mathematics
If $\alpha$ and $\beta$ are the roots of the equation $a^2+b x+c=0$, then the equation whose roots are $\alpha+\beta$ and $\frac{1}{\alpha}+\frac{1}{\beta}$ is
- A$a c x^2-(a b+b c) x+b^2=0$
- B$a c x^2+(a b+b c) x-b^2=0$
- Correct answerC$a c x^2+(a b+b c) x+b^2=0$
- D$a c x^2-(a b+b c) x-b^2=0$
Solution
Given : $\alpha$ and $\beta$ are roots of the equation
$a x^2+b x+c=0$
$\therefore \quad \alpha+\beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$
$\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac{b}{c}$ given by
$(x-(\alpha+\beta))\left(x-\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\right)=0$
$\Rightarrow\left(x+\frac{b}{a}\right)\left(x-\left(-\frac{b}{c}\right)\right)=0$
$\begin{aligned} & \Rightarrow \quad(a x+b)(c x+b)=0 \\ & \Rightarrow a c x^2+b c x+a b x+b^2=0 \\ & \Rightarrow a c x^2+x(a b+b c)+b^2=0 .\end{aligned}$
Hence, the correct answer is option: 'C'.