To find the sum of all real roots of the equation \(\sqrt{f(x) + 1} = 9701\), where \(f(x) = (x^2 + 3x)(x^2 + 3x + 2)\), we need to follow these steps:
Start with the equation:
\(\sqrt{f(x) + 1} = 9701\)
Squaring both sides to eliminate the square root gives:
\(f(x) + 1 = 9701^2\)
Calculate \(9701^2\):
\(9701^2 = 94109401\)
So, the equation becomes:
\((x^2 + 3x)(x^2 + 3x + 2) + 1 = 94109401\)
Let \(y = x^2 + 3x\). Then, \(f(x)\) simplifies to:
\(f(x) = y(y + 2) = y^2 + 2y\)
The equation becomes:
\(y^2 + 2y + 1 = 94109401\)
This simplifies to:
\(y^2 + 2y + 1 = 94109401\)
Rearranging gives:
\(y^2 + 2y + 1 - 94109401 = 0\)
Thus, the simplified quadratic equation is:
\(y^2 + 2y - 94109400 = 0\)
Now solve the quadratic equation:
\(y = \frac{-2 \pm \sqrt{(2)^2 + 4 \times 94109400}}{2}\)
\(= \frac{-2 \pm \sqrt{4 + 376437600}}{2}\)
\(= \frac{-2 \pm \sqrt{376437604}}{2}\)
Calculate the roots:
Given the equation roots are \(-3\), which corresponds to real solutions for this value of \(y\).
Thus, the sum of real roots of \(x\) itself from \(y \equiv x^2 + 3x\) equaling a quadratic with roots sum:
\(-b/a = -3/1 = -3\)
Conclusion:
The sum of all real roots of the equation is \(-3\).
This matches option: \(-3\), which is the correct answer.
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 no real roots. Therefore, the only real solutions are \(x = 97\) and \(x = -100\).
Step 4: Sum of all real roots. \[ \text{Sum of real roots} = 97 + (-100) = -3. \] \[ \boxed{-3} \]
For any natural number $k$, let $a_k = 3^k$. The smallest natural number $m$ for which \[ (a_1)^1 \times (a_2)^2 \times \dots \times (a_{20})^{20} \;<\; a_{21} \times a_{22} \times \dots \times a_{20+m} \] is: