A) positive
B) negative
C) of opposite signs
D) non real
Correct Answer: C
Solution :
We have\[\alpha +\beta =-\frac{p}{q},\,\,\,\,\alpha \beta =\frac{r}{p}\] Now the given equation \[\alpha {{(x-\beta )}^{2}}+\beta {{(x-\alpha )}^{2}}=0\] \[\Rightarrow \]\[(\alpha +\beta ){{x}^{2}}-4\alpha \beta x+\alpha \beta (\alpha +\beta )=0\] \[\Rightarrow \]\[\left( -\frac{q}{p} \right){{x}^{2}}-4\frac{r}{p}x+\frac{r}{p}\left( -\frac{q}{p} \right)=0\] \[\Rightarrow \]\[pq{{x}^{2}}+4prx+rq=0\] ???. (i) Since \[\alpha \] and \[\beta \] have opposite signs, therefore \[p\] and \[r\] must have opposite signs. \[\Rightarrow \]\[pq\]and \[rq\] must have opposite signs \[\Rightarrow \] roots of equation (i) have opposite signs
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