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 :

<p>$\alpha + \beta = \sqrt 2$, $\alpha \beta = \sqrt 6$</p> <p>$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$</p> <p>$= 2 + {{2 - 2\sqrt 6 } \over 6} = - a$</p> <p>$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}$$</p> <p>$= {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b$</p> <p>$\Rightarrow a + b = {{ - 5} \over 6}$</p> <p>So, equation is ${x^2} + {{17x} \over 6} + {7 \over 6} = 0$</p> <p>OR $6{x^2} + 17x + 7 = 0$</p> <p>Both roots of equation are $-$ve and distinct</p>