(N/A) Let the quadratic polynomial be $p(x) = ax^2 + bx + c$ and its zeroes be $\alpha$ and $\beta$.
Given,sum of zeroes $\alpha + \beta = \sqrt{2}$ and product of zeroes $\alpha \beta = -\frac{3}{2}$.
$A$ quadratic polynomial is given by $k[x^2 - (\alpha + \beta)x + \alpha \beta]$.
Substituting the values,we get $k[x^2 - \sqrt{2}x - \frac{3}{2}]$.
For $k=2$,the polynomial is $2x^2 - 2\sqrt{2}x - 3$.
To find the zeroes,set $2x^2 - 2\sqrt{2}x - 3 = 0$.
$2x^2 - 3\sqrt{2}x + \sqrt{2}x - 3 = 0$.
$\sqrt{2}x(\sqrt{2}x - 3) + 1(\sqrt{2}x - 3) = 0$.
$(\sqrt{2}x + 1)(\sqrt{2}x - 3) = 0$.
Thus,the zeroes are $x = -\frac{1}{\sqrt{2}}$ and $x = \frac{3}{\sqrt{2}}$.