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 $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :

If $$\alpha, \beta$$ are the roots of the equation

$$ x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0 $$,

then the equation, whose roots are $$\alpha+\frac{1}{\beta}$$ and $$\beta+\frac{1}{\alpha}$$, is :

The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:

Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :