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If $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$, and $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0 ; \mathrm{p}, \mathrm{q}$ and r being real numbers, then which of the following is not possible?
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Verified Answer
The correct answer is:
$\alpha=\beta=\mathrm{p}$
Given equation is $\mathrm{px}^{2}+\mathrm{qx}+\mathrm{r}=\mathrm{p}(\mathrm{x}-\alpha)(\mathrm{x}-\beta)$
$\Rightarrow p x^{2}+q x+r=p x^{2}-p(\alpha+\beta) x+\alpha \beta p$
$\Rightarrow \alpha \beta \mathrm{p}=\mathrm{r}$ and $\mathrm{q}=-(\alpha+\beta) \mathrm{p}$...(1)
Also given that $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0$
Putting value of $q$ and $r$ from $(1)$ $\Rightarrow p^{3}-p^{2}(\alpha+\beta)+\alpha \beta p=0$
$\Rightarrow \mathrm{p}^{2}-\mathrm{p}(\alpha+\beta)+\alpha \beta=0$
$\Rightarrow(p-\alpha)(p-\beta)=0 \Rightarrow \alpha=\beta=p$
$\Rightarrow p x^{2}+q x+r=p x^{2}-p(\alpha+\beta) x+\alpha \beta p$
$\Rightarrow \alpha \beta \mathrm{p}=\mathrm{r}$ and $\mathrm{q}=-(\alpha+\beta) \mathrm{p}$...(1)
Also given that $\mathrm{p}^{3}+\mathrm{pq}+\mathrm{r}=0$
Putting value of $q$ and $r$ from $(1)$ $\Rightarrow p^{3}-p^{2}(\alpha+\beta)+\alpha \beta p=0$
$\Rightarrow \mathrm{p}^{2}-\mathrm{p}(\alpha+\beta)+\alpha \beta=0$
$\Rightarrow(p-\alpha)(p-\beta)=0 \Rightarrow \alpha=\beta=p$
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