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(B) Given $\alpha + \beta = \sqrt{2}$ and $\alpha \beta = \sqrt{6}$.Let the roots of $x^{2}+ax+b=0$ be $y_1 = \frac{1}{\alpha^{2}}+1$ and $y_2 = \frac

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real and both negative

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(B) Given $\alpha + \beta = \sqrt{2}$ and $\alpha \beta = \sqrt{6}$.Let the roots of $x^{2}+ax+b=0$ be $y_1 = \frac{1}{\alpha^{2}}+1$ and $y_2 = \frac{1}{\beta^{2}}+1$.Sum of roots: $-a = y_1 + y_2 = \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 2 = \frac{(\alpha+\beta)^{2}-2\alpha \beta}{(\alpha \beta)^{2}} + 2 = \frac{2-2\sqrt{6}}{6} + 2 = \frac{1-\sqrt{6}}{3} + 2 = \frac{7-\sqrt{6}}{3}$.Product of roots: $b = y_1 y_2 = (\frac{1}{\alpha^{2}}+1)(\frac{1}{\beta^{2}}+1) = \frac{1}{(\alpha \beta)^{2}} + \frac{\alpha^{2}+\beta^{2}}{(\alpha \beta)^{2}} + 1 = \frac{1}{6} + \frac{2-2\sqrt{6}}{6} + 1 = \frac{1+2-2\sqrt{6}+6}{6} = \frac{9-2\sqrt{6}}{6}$.Now,consider the equation $x^{2}-(a+b-2)x+(a+b+2)=0$.$a+b = -\frac{7-\sqrt{6}}{3} + \frac{9-2\sqrt{6}}{6} = \frac{-14+2\sqrt{6}+9-2\sqrt{6}}{6} = -\frac{5}{6}$.The equation becomes $x^{2}-(-\frac{5}{6}-2)x+(-\frac{5}{6}+2) = 0$,which is $x^{2} + \frac{17}{6}x + \frac{7}{6} = 0$,or $6x^{2}+17x+7=0$.Roots are $x = \frac{-17 \pm \sqrt{289 - 168}}{12} = \frac{-17 \pm \sqrt{121}}{12} = \frac{-17 \pm 11}{12}$.$x_1 = -\frac{28}{12} = -\frac{7}{3}$ and $x_2 = -\frac{6}{12} = -\frac{1}{2}$.Both roots are real and negative.

Let $\alpha, \beta$ be the roots of the equation $x^{2}-\sqrt{2}x+\sqrt{6}=0$ and $\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$ be the roots of the equation $x^{2}+ax+b=0$. Then the roots of the equation $x^{2}-(a+b-2)x+(a+b+2)=0$ are...

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