Question

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

• A. \(6\)
• B. \(-3\)
• C. \(-6\)
• D. \(3\)

Hint:

When you see a product like $(x^2 + 3x)(x^2 + 3x + 2)$, try a substitution such as $t = x^2 + 3x$. Often, the expression simplifies to a perfect square (like $(t+1)^2$ here), which makes equations with square roots much easier to solve.

Answer 1

The Correct Option is B

Approach: Skip the substitution and attack $f(x)+1$ as an algebraic identity. Group the consecutive parts so the product of two "adjacent" expressions plus $1$ becomes a perfect square directly.

Step 1: Spot the identity. Write $a = x^2+3x$. Then $f(x)+1 = a(a+2)+1 = a^2+2a+1 = (a+1)^2 = (x^2+3x+1)^2.$
So the equation $\sqrt{f(x)+1}=9701$ becomes $|x^2+3x+1| = 9701.$

Step 2: Drop the modulus. Either
(i) $x^2+3x+1 = 9701 \Rightarrow x^2+3x-9700 = 0$, or
(ii) $x^2+3x+1 = -9701 \Rightarrow x^2+3x+9702 = 0.$

Step 3: Test for real roots via discriminant.
(i) $\Delta = 9 + 4(9700) = 38809 = 197^2 > 0 \Rightarrow$ two real roots, summing to $-3$ (Vieta).
(ii) $\Delta = 9 - 4(9702) = -38799 < 0 \Rightarrow$ no real roots, contributes nothing.

Step 4: Sum of all real roots $= -3.$

\[ \text{Sum} = -3 \]