Given the equation 2x²−3x−6= are alpha and beta find (alpha square+1/beta square) are beta square+1/alpha square

For a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product of the roots is $\alpha \beta = \frac{c}{a}$.
Here, $a=2$, $b=-3$, $c=-6$.
Sum of roots: $\alpha + \beta = -\frac{-3}{2} = \frac{3}{2}$
Product of roots: $\alpha \beta = \frac{-6}{2} = -3$

Let $\alpha = \frac{3 + \sqrt{57}}{4}$ and $\beta = \frac{3 - \sqrt{57}}{4}$.

$\alpha^2 + \frac{1}{\beta^2}$.
We know $\alpha\beta = -3$, so $\frac{1}{\beta} = -\frac{\alpha}{3}$.
Therefore, $\frac{1}{\beta^2} = \left(-\frac{\alpha}{3}\right)^2 = \frac{\alpha^2}{9}$.
Substitute this into the expression:

Now substitute the value of $\alpha^2$:

$\beta^2 + \frac{1}{\alpha^2}$.
Similarly, $\frac{1}{\alpha} = -\frac{\beta}{3}$.
Therefore, $\frac{1}{\alpha^2} = \left(-\frac{\beta}{3}\right)^2 = \frac{\beta^2}{9}$.
Substitute this into the expression:

Now substitute the value of $\beta^2$:

The values are: