The condition that $x^3-p x^2+q x-r=0$ may have two of its roots equal to each other but of opposite sign is
The condition that $x^3-p x^2+q x-r=0$ may have two of its roots equal to each other but of opposite sign is
- r = pq
- $r=2 p^3+p q$
- $r=p^2 q$
- $r=p^2 q^2$
Solution
Given equation, $x^3-p x^2+q x-r=0 ...(i)\ldots$
Let $\alpha, \beta$ and $\gamma$ are the roots of Eq. (i) Hence,
$
\begin{array}{r}
\alpha+\beta+\gamma=\frac{-(-p)}{1}=p ...(ii)\\
\alpha \beta+\beta \gamma+\gamma \alpha=q ...(iii)\\
\alpha \beta \gamma=-(-r)=r..._(iv)
\end{array}
$
Given that, two roots are equal and opposite in sign.
Let $\alpha=-\beta$
From Eq. (ii), we get, $\alpha+\beta+\gamma=p$
$
\begin{aligned}
-\beta+\beta+\gamma & =p \\
\gamma & =p...(v)
\end{aligned}
$
From Eq. (iv), we get, $\alpha \beta \gamma=r$
$
\Rightarrow
$
$
\begin{aligned}
-\beta^2 \gamma & =r \\
-\beta^2 p & =r \\
\beta^2 & =\frac{-r}{p}
\end{aligned}
$
From Eq. (iii), we get
$
\begin{aligned}
\alpha \beta+\beta \gamma+\gamma \alpha & =q \\
-\beta^2+\beta p+(-\beta) p & =q \Rightarrow \frac{r}{p}+0=q
\end{aligned}
$
$
\Rightarrow \quad r=p q
$
Asked in: AP EAMCET 2021 (25 Aug Shift 2)
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