A) \[4{{x}^{2}}+7x+16=0\]
B) \[4{{x}^{2}}+7x+6=0\]
C) \[4{{x}^{2}}+7x+1=0\]
D) \[4{{x}^{2}}-7x+16=0\]
Correct Answer: A
Solution :
\[\alpha +\beta =\frac{3}{2}\] and \[\alpha \beta =2\] \[{{\alpha }^{2}}+{{\beta }^{2}}={{(\alpha +\beta )}^{2}}-2\alpha \beta =\frac{9}{4}-4=-\frac{7}{4}\] Hence required equation \[{{x}^{2}}-({{\alpha }^{2}}+{{\beta }^{2}})x+{{\alpha }^{2}}{{\beta }^{2}}=0\] Þ \[{{x}^{2}}+\frac{7}{4}x+4=0\]Þ \[4{{x}^{2}}+7x+16=0\]
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