For the given sum and product of zeroes,find a quadratic polynomial. Also,find the zeroes of this polynomial by factorisation:
Sum of zeroes = $\frac{-3}{2 \sqrt{5}}$,Product of zeroes = $-\frac{1}{2}$

(N/A) Given that,Sum of zeroes $(S)$ = $-\frac{3}{2 \sqrt{5}}$ and Product of zeroes $(P)$ = $-\frac{1}{2}$.
The general form of a quadratic polynomial is $f(x) = k(x^2 - Sx + P)$,where $k$ is a constant.
Substituting the values,$f(x) = x^2 - (-\frac{3}{2 \sqrt{5}})x + (-\frac{1}{2}) = x^2 + \frac{3}{2 \sqrt{5}}x - \frac{1}{2}$.
To simplify,we can multiply by $2\sqrt{5}$ to get the polynomial $2\sqrt{5}x^2 + 3x - \sqrt{5}$.
Now,factorising $2\sqrt{5}x^2 + 3x - \sqrt{5}$:
$= 2\sqrt{5}x^2 + 5x - 2x - \sqrt{5}$
$= \sqrt{5}x(2x + \sqrt{5}) - 1(2x + \sqrt{5})$
$= (2x + \sqrt{5})(\sqrt{5}x - 1)$
Setting $f(x) = 0$,we get $2x + \sqrt{5} = 0$ or $\sqrt{5}x - 1 = 0$.
Therefore,the zeroes are $x = -\frac{\sqrt{5}}{2}$ and $x = \frac{1}{\sqrt{5}}$.