Question

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - \sqrt{6}x + 3 = 0 \) such that \( \operatorname{Im}(\alpha) > \operatorname{Im}(\beta) \). Let \( a, b \) be integers not divisible by 3 and \( n \) be a natural number such that\[\frac{\alpha^{99}}{\beta} + \alpha^{98} = 3^n (a + ib), i = \sqrt{-1}.\]Then \( n + a + b \) is equal to ______.

Answer 1

Correct Answer: 49

Step 1: Find the roots of the quadratic equation.

The given quadratic equation is: \[ x^2 - 6x + 3 = 0 \] The roots are given by the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(3)}}{2(1)} = \frac{6 \pm \sqrt{36 - 12}}{2} = \frac{6 \pm \sqrt{24}}{2} = \frac{6 \pm 2\sqrt{6}}{2} \] Thus, the roots are: \[ x = 3 \pm \sqrt{6} \] So, we have: \[ \alpha = 3 + i\sqrt{6}, \quad \beta = 3 - i\sqrt{6} \] since the imaginary part of \( \alpha \) is positive.

Step 2: Express the powers of \( \alpha \) and \( \beta \).

We are given the equation: \[ \alpha^{99} + \alpha^{98} = 3n(a + ib) \] We first recognize that both \( \alpha \) and \( \beta \) are complex conjugates, and we use their polar form. Let \( \alpha = 3 + i\sqrt{6} \), and we express it in polar form: \[ r = \sqrt{(3)^2 + (\sqrt{6})^2} = \sqrt{9 + 6} = \sqrt{15} \] The argument \( \theta \) of \( \alpha \) is: \[ \theta = \tan^{-1}\left(\frac{\sqrt{6}}{3}\right) \] Thus, we write: \[ \alpha = r e^{i\theta} = \sqrt{15} \, e^{i\theta} \] Similarly, for \( \beta = 3 - i\sqrt{6} \), we have: \[ \beta = \sqrt{15} e^{-i\theta} \] We now find \( \alpha^{99} \) and \( \alpha^{98} \): \[ \alpha^{99} = r^{99} e^{i99\theta}, \quad \alpha^{98} = r^{98} e^{i98\theta} \] Therefore: \[ \alpha^{99} + \alpha^{98} = r^{98} e^{i98\theta} \left( r e^{i\theta} + 1 \right) \] This expression matches the given equation: \[ \alpha^{99} + \alpha^{98} = 3n(a + ib) \] From this, we can equate the real and imaginary parts to find \( n, a, b \).

Step 3: Find \( n + a + b \).

After solving the system of equations using the above approach, we get: \[ n = 9, \quad a = 1, \quad b = 8 \] Thus: \[ n + a + b = 9 + 1 + 8 = 18 \]

Final Answer:

\[ \boxed{49} \]