If \( f ( x ) = \left( x ^ { 2 } + 3 x \right) \left( x ^ { 2 } + 3 x + 2 \right) \), then the sum of all real roots of the equation \( \sqrt { f ( x ) + 1 } = 9701 \), is
Answer: B
Key idea: Substitute \( y = x^2 + 3x \); then \( f(x) + 1 \) becomes a perfect square.
1. \( f(x) + 1 = y(y + 2) + 1 = (y + 1)^2 \), so \( \sqrt{f(x) + 1} = |y + 1| = 9701 \).
2. Case \( y + 1 = 9701 \): \( x^2 + 3x - 9700 = 0 \). Its discriminant is positive, so it has two real roots with sum \( -3 \).
3. Case \( y + 1 = -9701 \): \( x^2 + 3x + 9702 = 0 \). Discriminant \( 9 - 4 \times 9702 < 0 \), so no real roots.
So the sum of all real roots is \( -3 \).
Tip: \( x^2 + 3x \ge -\frac{9}{4} \) for every real \( x \), so \( y = -9702 \) is impossible without computing anything.
Hence, the answer is (B) \( -3 \).