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

Step 1: Simplify the expression inside the square root. Given \[ f(x) = (x^2 + 3x)(x^2 + 3x + 2). \] Let \[ t = x^2 + 3x. \] Then \[ f(x) = t(t+2) = t^2 + 2t, \] so \[ f(x) + 1 = t^2 + 2t + 1 = (t+1)^2. \] The given equation is: \[ \sqrt{f(x) + 1} = 9701 \quad \Rightarrow \quad \sqrt{(t+1)^2} = 9701. \] Since the square root is non-negative, \[ |t+1| = 9701 \quad \Rightarrow \quad t+1 = 9701 \ \text{or}\ t+1 = -9701. \] Thus, \[ t = 9700 \quad \text{or} \quad t = -9702. \] Recall that \(t = x^2 + 3x\), so we get two quadratic equations: \[ x^2 + 3x = 9700 \quad \Rightarrow \quad x^2 + 3x - 9700 = 0, \] \[ x^2 + 3x = -9702 \quad \Rightarrow \quad x^2 + 3x + 9702 = 0. \]
Step 2: Solve the first quadratic. For \[ x^2 + 3x - 9700 = 0, \] the discriminant is \[ \Delta_1 = 3^2 + 4 \cdot 9700 = 9 + 38800 = 38809. \] \[ \sqrt{38809} = 197, \] so the roots are \[ x = \frac{-3 \pm 197}{2}. \] Thus: \[ x_1 = \frac{-3 + 197}{2} = \frac{194}{2} = 97,\quad x_2 = \frac{-3 - 197}{2} = \frac{-200}{2} = -100. \]
Step 3: Check the second quadratic for real roots. For \[ x^2 + 3x + 9702 = 0, \] the discriminant is \[ \Delta_2 = 3^2 - 4 \cdot 9702 = 9 - 38808 = -38799<0, \] so this quadratic has \emph{no real roots}. Therefore, the only real solutions are \(x = 97\) and \(x = -100\).
Step 4: Sum of all real roots. \[ \text{Sum of real roots} = 97 + (-100) = -3. \] \[ \boxed{-3} \]