Let α, β be the roots of the equation x ^ 2 - 2 x + 6 = 0 and 1 α ^...

Given $\alpha, \beta$ be the roots of the equation ${x}^{2} - \sqrt{2}x + \sqrt{6} = 0$

So sum of roots will be $\alpha + \beta = \sqrt{2}$ and product of roots will be $\alpha\beta = \sqrt{6}$

And also given $\dfrac{1}{{\alpha}^{2}} + 1$ and $\dfrac{1}{{\beta}^{2}} + 1$ are roots of ${x}^{2} + ax + b = 0$

So sum of roots will be $- a = \dfrac{1}{{\alpha}^{2}} + 1 + \dfrac{1}{{\beta}^{2}} + 1$

$⇒ a = \dfrac{- 1}{{\alpha}^{2}} - \dfrac{1}{{\beta}^{2}} - 2 . . . . . \left(1\right)$

And similarly product of roots will be,

$b = \dfrac{1}{{\alpha}^{2}} + \dfrac{1}{{\beta}^{2}} + 1 + \dfrac{1}{{\alpha}^{2}{\beta}^{2}} . . . . \left(2\right)$

Now adding equation \left(1\right) \text{ and } \left(2\right)$we get,</p><p>$a + b = \dfrac{1}{{\left(\alpha\beta\right)}^{2}} - 1 = \dfrac{1}{6} - 1 = - \dfrac{5}{6}${as \alpha\beta = \sqrt{6}}$

Now putting the value of $a + b$ in ${x}^{2} - \left(a + b - 2\right)x + \left(a + b + 2\right) = 0$

$⇒ {x}^{2} - \left(- \dfrac{5}{6} - 2\right)x + \left(2 - \dfrac{5}{6}\right) = 0$

$⇒ 6{x}^{2} + 17x + 7 = 0$

⇒ x = - \dfrac{7}{3}, x = - \dfrac{1}{2} are the roots, both roots are real and negative.