A) -2
B) -1
C) 1
D) 2
Correct Answer: C
Solution :
Given that \[\alpha \]and \[\beta \]be the roots of \[{{x}^{2}}+x+1=0\], so \[\alpha +\beta =-1\]and \[\alpha \beta =1\] Again \[\frac{\alpha }{\beta }\]and \[\frac{\beta }{\alpha }\]are the roots of \[{{x}^{2}}+px+q=0,\] so \[\frac{\alpha }{\beta }+\frac{\beta }{\alpha }=-p\] Þ \[-p=\frac{{{\alpha }^{2}}+{{\beta }^{2}}}{\alpha \beta }\] Þ \[-p=\frac{{{(\alpha +\beta )}^{2}}-2\alpha \beta }{\alpha \beta }=\frac{1-2}{1}\Rightarrow p=1\]
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