If α and β are the roots of the quadratic equation x^2 + x – 1 = 0, then what is the equation whose roots are α^5 and β^5?

Introduction / Context:
This problem is about transforming roots of a quadratic equation. We are given the roots α and β of a basic quadratic and are asked to find a new quadratic whose roots are α^5 and β^5. This type of question tests knowledge of relationships between roots and coefficients and the use of algebraic manipulation.

Given Data / Assumptions:

• α and β are the roots of x^2 + x - 1 = 0.
• We must find the quadratic equation whose roots are α^5 and β^5.
• All coefficients are real.

Concept / Approach:
For a quadratic with roots r1 and r2, the sum of roots is r1 + r2 and the product of roots is r1 r2. For a new quadratic with roots r1^5 and r2^5, the sum of roots is r1^5 + r2^5 and the product is (r1 r2)^5. Using the original equation, we know α + β and αβ. We compute α^5 + β^5 using recurrence relations or algebraic manipulation and then construct the new quadratic as x^2 - (sum of new roots) x + (product of new roots) = 0.

Step-by-Step Solution:
From x^2 + x - 1 = 0, for roots α and β, we have α + β = -1 and αβ = -1. Let S1 = α + β = -1 and P = αβ = -1. We want S5 = α^5 + β^5 and P5 = (αβ)^5. Compute P5 first: P5 = (αβ)^5 = (-1)^5 = -1. To find S5, we can use the recurrence for powers of roots of a quadratic: for n ≥ 2, S_n = -S_{n-1} + S_{n-2}, derived from the relation α^2 = -α + 1 and β^2 = -β + 1. First compute S0 and S2. We have S0 = α^0 + β^0 = 1 + 1 = 2. S1 = α + β = -1. Using the recurrence S2 = -S1 + S0 = -(-1) + 2 = 3. Next, S3 = -S2 + S1 = -3 + (-1) = -4. Then S4 = -S3 + S2 = -(-4) + 3 = 7. Finally, S5 = -S4 + S3 = -7 + (-4) = -11. Thus α^5 + β^5 = S5 = -11 and (αβ)^5 = -1. The quadratic equation with roots α^5 and β^5 is x^2 - (sum of roots) x + product of roots = 0. Substitute: x^2 - ( -11 ) x + ( -1 ) = 0, which simplifies to x^2 + 11x - 1 = 0.

Verification / Alternative check:
We can verify by finding numerical approximations of α and β, raising them to the fifth power, and then checking whether x^2 + 11x - 1 = 0 has those approximate values as roots. Substituting α^5 and β^5 into x^2 + 11x - 1 confirms that the left side becomes zero within numerical precision.

Why Other Options Are Wrong:
x^2 - 7x - 1 = 0 and x^2 + 7x - 1 = 0: These have sums of roots equal to 7 and -7, which do not match S5 = -11.
x^2 - 11x - 1 = 0: This has sum of roots equal to 11 instead of -11, reversing the sign.
x^2 + 5x - 1 = 0: The sum of roots here is -5, which again does not match S5.

Common Pitfalls:
Learners often forget the recurrence relation for sums of powers of roots and attempt to expand α^5 and β^5 directly, which is very tedious. Another mistake is misapplying the sign in the relation between coefficients and sums of roots. Using systematic recurrence and checking signs carefully solves these issues.

Final Answer:
The equation whose roots are α^5 and β^5 is x^2 + 11x - 1 = 0.