Question

Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to:

Hint:

In problems involving powers of roots of a quadratic equation, use recurrence relations to compute higher powers and solve for desired expressions.

Answer 1

Correct Answer: 31

Let \( \alpha \) and \( \beta \) be the roots of the quadratic equation \( x^2 - ax - b = 0 \). Define \( P_n = \alpha^n - \beta^n \). The objective is to determine \( |\alpha^4 + \beta^4| \).

From Vieta's formulas:

• Sum of roots: \( \alpha + \beta = a \)
• Product of roots: \( \alpha \beta = b \)

The sequence \( P_n \) satisfies the recurrence relation:

• \( P_n = (\alpha + \beta)P_{n-1} - \alpha\beta P_{n-2} \)

Given values:

• \(P_3 = -5\sqrt{7}\)
• \(P_4 = -3\sqrt{7}\)
• \(P_5 = 11\sqrt{7}\)
• \(P_6 = 45\sqrt{7}\)

Using the recurrence relation with \( P_3 \) and \( P_4 \):

\( P_5 = (\alpha + \beta)P_4 - \alpha\beta P_3 \)

\( 11\sqrt{7} = a(-3\sqrt{7}) - b(-5\sqrt{7}) \)

This simplifies to the equation:

\( 11 = -3a + 5b \) (i)

Using the recurrence relation with \( P_4 \) and \( P_5 \):

\( P_6 = (\alpha + \beta)P_5 - \alpha\beta P_4 \)

\( 45\sqrt{7} = a(11\sqrt{7}) - b(-3\sqrt{7}) \)

This simplifies to the equation:

\( 45 = 11a + 3b \) (ii)

Solving the system of linear equations:

\( -3a + 5b = 11 \) (i)

\( 11a + 3b = 45 \) (ii)

Multiply equation (i) by 3 and equation (ii) by 5, then add:

\( 3 \times (-3a + 5b) = 3 \times 11 \Rightarrow -9a + 15b = 33 \)

\( 5 \times (11a + 3b) = 5 \times 45 \Rightarrow 55a + 15b = 225 \)

Subtracting the first new equation from the second yields:

\( (55a + 15b) - (-9a + 15b) = 225 - 33 \)

\( 64a = 192 \Rightarrow a = 3 \)

Substitute \( a = 3 \) into equation (i):

\( -3(3) + 5b = 11 \Rightarrow -9 + 5b = 11 \Rightarrow 5b = 20 \Rightarrow b = 4 \)

Now, calculate \( \alpha^4 + \beta^4 \). We know that:

\( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta \)

\( \alpha^2 + \beta^2 = a^2 - 2b = 3^2 - 2(4) = 9 - 8 = 1 \)

Then:

\( \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2 \)

\( \alpha^4 + \beta^4 = (1)^2 - 2(4)^2 = 1 - 2(16) = 1 - 32 = -31 \)

Therefore, \( |\alpha^4 + \beta^4| = |-31| = 31 \).

Final Answer: \(|\alpha^4 + \beta^4| = 31\)