Sum of smallest roots 
If 2 and -3 are the roots of a biquadratic equation, then the sum of the two smallest
  roots of this equation is: $\{-1, -3, -5, \text{cannot be determined} \}$ those are the options.

The answer key says C = $-5$.
But, 
$P(x) = (x-2)(x+3)(f(x))$ where $f(x)$ has degree two.
It is possible to have: $P(x) = (x-2)(x+3)(x+2)(x-3)$ which gives $-5$ but there are other options such as:
$P(x) = (x-2)(x+3)(x+1999919199919912919199191)(x+1919359237597373792397)$
What is it then?

Biquartic functions have the form 
$$
Q(z)=a_4z^4+a_2z^2+a_0
$$
Each solution $z_\pm(\neq 0)$ of $Q(z)=0$ splits in two: $\pm\sqrt{z_\pm}$.
So when $2$ and $-3$ are root, $-2$ and $3$ are too, so the sum of the smallest is $-5$.
