Sum of Roots of Radical Function

Let $y = x^2 + 3x$. Then $f(x) = y(y + 2) = y^2 + 2y$. $f(x) + 1 = y^2 + 2y + 1 = (y + 1)^2 = (x^2 + 3x + 1)^2$. Therefore, $\sqrt{f(x) + 1} = |x^2 + 3x + 1| = 9701$. Since $9701 > 0$, we have two quadratic equations:

1. $x^2 + 3x + 1 = 9701 \implies x^2 + 3x - 9700 = 0$.
2. $x^2 + 3x + 1 = -9701 \implies x^2 + 3x + 9702 = 0$. For equation 1: $D = 9 - 4(-9700) > 0$, real roots exist. Sum of roots = $-3$. For equation 2: $D = 9 - 4(9702) < 0$, no real roots. Thus, the sum of all real roots is $-3$.