Key Concepts and Formulas

• Quadratic Formula: For a quadratic equation of the form $ax^2 + bx + c = 0$, the roots are given by $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$:
  • Sum of roots: $\alpha + \beta = -\frac{b}{a}$
  • Product of roots: $\alpha \beta = \frac{c}{a}$
• Properties of Exponents: $a^{m+n} = a^m \cdot a^n$ and $a^{mn} = (a^m)^n$

Step-by-Step Solution

Step 1: Find the roots of the equation. We are given the quadratic equation $x^2 - x + 1 = 0$. To find the roots, we use the quadratic formula with $a=1$, $b=-1$, and $c=1$: $x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(1)}}{2(1)}$ $x = \frac{1 \pm \sqrt{1 - 4}}{2}$ $x = \frac{1 \pm \sqrt{-3}}{2}$ $x = \frac{1 \pm i\sqrt{3}}{2}$ So, the roots are $\alpha = \frac{1 + i\sqrt{3}}{2}$ and $\beta = \frac{1 - i\sqrt{3}}{2}$.

Step 2: Identify a key property of these roots. Consider the given equation $x^2 - x + 1 = 0$. Multiplying this equation by $(x+1)$ gives: $(x+1)(x^2 - x + 1) = 0$ Expanding this yields the sum of cubes formula: $x^3 + 1^3 = 0$ $x^3 + 1 = 0$ $x^3 = -1$ This means that any root of $x^2 - x + 1 = 0$ must satisfy $x^3 = -1$. Therefore, we have $\alpha^3 = -1$ and $\beta^3 = -1$. This is a crucial step to simplifying high powers of $\alpha$ and $\beta$.

Step 3: Simplify the exponents of the roots. We need to calculate $\alpha^{2009} + \beta^{2009}$. To simplify the exponent 2009, we divide it by 3: $2009 \div 3 = 669 \text{ with a remainder of } 2$ So, we can write $2009 = 3 \times 669 + 2$. Applying this to $\alpha^{2009}$: $\alpha^{2009} = \alpha^{3 \times 669 + 2}$ Using the exponent rules $a^{m+n} = a^m \cdot a^n$ and $a^{mn} = (a^m)^n$: $\alpha^{2009} = (\alpha^3)^{669} \cdot \alpha^2$ Since $\alpha^3 = -1$: $\alpha^{2009} = (-1)^{669} \cdot \alpha^2$ As 669 is an odd number, $(-1)^{669} = -1$. Therefore, $\alpha^{2009} = -\alpha^2$ Similarly, for $\beta^{2009}$: $\beta^{2009} = (\beta^3)^{669} \cdot \beta^2 = (-1)^{669} \cdot \beta^2 = -\beta^2$

Step 4: Calculate the sum of the simplified powers. Now we substitute these simplified forms back into the expression we want to find: $\alpha^{2009} + \beta^{2009} = -\alpha^2 - \beta^2$ $\alpha^{2009} + \beta^{2009} = -(\alpha^2 + \beta^2)$

Step 5: Find the value of $\alpha^2 + \beta^2$ using Vieta's formulas. From Vieta's formulas applied to $x^2 - x + 1 = 0$:

• Sum of roots: $\alpha + \beta = -\frac{-1}{1} = 1$.
• Product of roots: $\alpha \beta = \frac{1}{1} = 1$.

We need to find $\alpha^2 + \beta^2$. We can use the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$. Rearranging this identity to solve for $\alpha^2 + \beta^2$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ Now, substitute the values from Vieta's formulas: $\alpha^2 + \beta^2 = (1)^2 - 2(1)$ $\alpha^2 + \beta^2 = 1 - 2$ $\alpha^2 + \beta^2 = -1$

Step 6: Perform the final calculation. Substitute the value of $\alpha^2 + \beta^2$ from Step 5 into the expression from Step 4: $\alpha^{2009} + \beta^{2009} = -(\alpha^2 + \beta^2)$ $\alpha^{2009} + \beta^{2009} = -(-1)$ $\alpha^{2009} + \beta^{2009} = 1$

Common Mistakes & Tips

• Incorrectly simplifying exponents: Double-check the remainder when dividing exponents.
• Errors with powers of negative numbers: Remember that $(-1)^{\text{odd number}} = -1$ and $(-1)^{\text{even number}} = 1$.
• Calculation errors in Vieta's formulas: Ensure the sum and product of roots are correctly derived from the coefficients of the quadratic equation.

Summary

The problem is efficiently solved by recognizing that the roots of $x^2 - x + 1 = 0$ satisfy the property $x^3 = -1$. This allows us to reduce the high exponent 2009 to a small exponent 2. The problem then simplifies to calculating $-(\alpha^2 + \beta^2)$. Vieta's formulas provide a straightforward way to find $\alpha + \beta = 1$ and $\alpha \beta = 1$, enabling us to compute $\alpha^2 + \beta^2 = -1$. Combining these results, we find $\alpha^{2009} + \beta^{2009} = -(-1) = 1$.

Final Answer

The final answer is $\boxed{1}$, which corresponds to option (B).