A) \[3\sqrt{2}{{x}^{2}}+7x+3\]
B) \[\sqrt{6}{{x}^{2}}+5x+\sqrt{6}\]
C) \[\sqrt{6}{{x}^{2}}+5x+\sqrt{6}\]
D) \[3{{x}^{2}}+\frac{5}{2\sqrt{3}}x-\frac{1}{6}\]
Correct Answer: B
Solution :
(b): Simple method \[\alpha =\frac{-\sqrt{3}}{\sqrt{2}}\therefore \beta =\frac{-\sqrt{2}}{\sqrt{3}}\] \[\therefore {{x}^{2}}+\left( \frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{2}}{\sqrt{3}} \right)x+1=0\] Elegant solution: To provide elegance to our approach, we say that the given quadratic polynomial will have \[\alpha \] & \[\frac{1}{\alpha }\] as roots. (This straightaway removes one variable as \[\beta \]is written as \[\frac{1}{\alpha }\]) \[\therefore p(x)={{x}^{2}}-\left( \alpha +\frac{1}{\alpha } \right)x+1\] \[={{x}^{2}}-\left( \frac{{{\alpha }^{2}}+1}{\alpha } \right)+1\] \[\alpha =\frac{-\sqrt{3}}{\sqrt{2}}\therefore {{\alpha }^{2}}=\frac{3}{2}\therefore {{\alpha }^{2}}+1=\frac{5}{2}\]\[\therefore \frac{{{\alpha }^{2}}+1}{\alpha }=\frac{5\sqrt{2}}{-\sqrt{3}/\sqrt{2}}=-\frac{5}{\sqrt{6}}\] \[\therefore p(x)={{x}^{2}}+\frac{5}{\sqrt{6}}x+1\] or a multiple of \[\sqrt{6}{{x}^{2}}+5x+\sqrt{6}\]
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