⚡ AZUCATION SHORTCUT:
Let $t = x^2 + 3x$. The equation becomes $\sqrt{t(t+2) + 1} = 9701 \implies \sqrt{(t+1)^2} = 9701$.
So, $|t+1| = 9701$. For the case $x^2 + 3x - 9700 = 0$, the sum of roots is $-b/a = \mathbf{-3}$.
The other case yields non-real roots.

Step 1: Simplify the expression $f(x)$
Let $x^2 + 3x = t$.
Then $f(x) = t(t + 2) = t^2 + 2t$.

Step 2: Solve the radical equation
Substitute $f(x)$ into the equation:
$\sqrt{(t^2 + 2t) + 1} = 9701$
$\sqrt{(t + 1)^2} = 9701$
$|t + 1| = 9701$

Step 3: Analyze the cases for $t$
Case 1: $t + 1 = 9701 \implies t = 9700$
$x^2 + 3x = 9700 \implies x^2 + 3x - 9700 = 0$
Discriminant $D = 3^2 - 4(1)(-9700) > 0$. Real roots exist.
Sum of roots ($S_1$) $= -3/1 = -3$.

Case 2: $t + 1 = -9701 \implies t = -9702$
$x^2 + 3x = -9702 \implies x^2 + 3x + 9702 = 0$
Discriminant $D = 3^2 - 4(1)(9702) < 0$. No real roots exist.

Step 4: Final Conclusion
The only real roots come from Case 1.
Sum of all real roots $= \mathbf{-3}$.

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