If $$f(x)= (x^{2} + 3x)(x^{2}+ 3x+2)$$ then the sum of all real roots of the equation $$\sqrt{f(x)+1}= 9701$$, is

Let $$(x^2+3x)$$ be equal to $$k$$. We have, 

$$f(x)= k(k+2) = k^2+2k$$ 

Therefore, $$\sqrt{f(x)+1} = \sqrt{k^2+2k+1} = \sqrt{(k+1)^2} = k+1 = 9701$$

We get $$k=9700$$.

Thus, $$x^2+3x=9700$$ or $$x^2+3x-9700 = 0$$

Since $$x$$ is real, the discriminant of the above quadratic has to be greater than or equal to zero.

We find that $$3^2 + 4*9700 \geq 0$$ and therefore the quadratic has real roots.

The sum of the roots will be $$-\dfrac{b}{a} = -\dfrac{3}{1} = -3$$

Option D is the correct answer.