Theory Of Equations Ques 18
- Let $p$ and $q$ be real numbers such that $p \neq 0, p^{3} \neq q$ and $p^{3} \neq-q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^{3}+\beta^{3}=q$, then a quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is
(2010)
(a) $\left(p^{3}+q\right) x^{2}-\left(p^{3}+2 q\right) x+\left(p^{3}+q\right)=0$
(b) $\left(p^{3}+q\right) x^{2}-\left(p^{3}-2 q\right) x+\left(p^{3}+q\right)=0$
(c) $\left(p^{3}-q\right) x^{2}-\left(5 p^{3}-2 q\right) x+\left(p^{3}-q\right)=0$
(d) $\left(p^{3}-q\right) x^{2}-\left(5 p^{3}+2 q\right) x+\left(p^{3}-q\right)=0$
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Answer:
Correct Answer: 18.(b)
Solution:
- Sum of roots $=\frac{\alpha^{2}+\beta^{2}}{\alpha \beta}$ and product $=1$
Given, $\alpha+\beta=-p$ and $\alpha^{3}+\beta^{3}=q$
$ \begin{aligned} & \Rightarrow \quad(\alpha+\beta)\left(\alpha^{2}-\alpha \beta+\beta^{2}\right)=q \\ & \therefore \quad \alpha^{2}+\beta^{2}-\alpha \beta=\frac{-q}{p} ……(i) \end{aligned} $
and
$ (\alpha+\beta)^{2}=p^{2} $
$ \alpha^{2}+\beta^{2}+2 \alpha \beta=p^{2} ……(ii) $
From Eqs. (i) and (ii), we get
$ \alpha^{2}+\beta^{2}=\frac{p^{3}-2 q}{3 p} \quad \text { and } \quad \alpha \beta=\frac{p^{3}+q}{3 p} $
$\therefore$ Required equation is, $x^{2}-\frac{\left(p^{3}-2 q\right) x}{\left(p^{3}+q\right)}+1=0$
$\Rightarrow\left(p^{3}+q\right) x^{2}-\left(p^{3}-2 q\right) x+\left(p^{3}+q\right)=0$
BETA
