Problem 154 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(a\) and \(\beta\) are the roots of the equation \(x^{2}-x+1=0\) then the value of \(\alpha^{2009}+\beta^{2009}=\) [2010] (A) \(-1\) (B) 1 (C) 2 (D) \(-2\)

Step-by-Step Solution

Answer

(A) -1

Step 1: Identify the Polynomial Roots

The given quadratic equation is \(x^2 - x + 1 = 0\). Let's identify its roots, \(\alpha\) and \(\beta\). We solve using the quadratic formula, \(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), where \(a = 1, b = -1, c = 1\).

Step 2: Apply the Quadratic Formula

Apply the quadratic formula: \[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2} \].

Step 3: Simplify the Roots

The roots are complex numbers: \[ \alpha = \frac{1 + i\sqrt{3}}{2}, \beta = \frac{1 - i\sqrt{3}}{2} \]. These roots, \(\alpha\) and \(\beta\), are complex conjugates.

Step 4: Use Properties of Powers of Roots

Notice that \(\alpha\) and \(\beta\) are complex conjugates. In such cases, powers of complex conjugate roots can be simplified using identities. Particularly, for polynomials \(\alpha^n + \beta^n\), there are well-known formulas for repetitive cycles.

Step 5: Compute Powers

Analyze the cycling behavior by starting with small powers. Compute \(\alpha^0 + \beta^0 = 2\), \(\alpha^1 + \beta^1 = 1\), \(\alpha^2 + \beta^2 = -1\), and \(\alpha^3 + \beta^3 = -1\). You'll notice a repeating pattern every 3: 2, 1, -1, -1.

Step 6: Identify the Cycle for n = 2009

Recognize that \(\alpha^n + \beta^n\) follows: 2, 1, -1, -1 in a cycle of length 3. Determine the equivalent position for \(n = 2009\): 2009 mod 3 = 2. From the pattern, when \(n \equiv 2\), the result is -1.

Key Concepts

Quadratic Equations

A quadratic equation is a second-degree polynomial of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). It represents a parabola when graphed on the Cartesian plane. Quadratics can have two distinct real roots, one repeated real root, or two complex roots.
To find these roots, we often use the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] This formula gives us possible solutions for \(x\). The expression \(b^2 - 4ac\) is known as the discriminant.

• If the discriminant is positive, we have two real and distinct roots.
• If it is zero, we have exactly one real root, also called a repeated or double root.
• If it is negative, the equation has two complex roots that are conjugates of each other.

In our problem, since the discriminant is negative, the quadratic equation has complex roots.

Roots of Equations

The roots of an equation are the values of the variable that satisfy the equation, making it true. For quadratic equations, these roots can be real or complex. In our exercise, the roots \( \alpha \) and \( \beta \) of the equation \(x^2 - x + 1 = 0\) are complex.
Complex roots often occur in conjugate pairs when the polynomial has real coefficients. Complex conjugate roots mean if one root is \(a + bi\), the other will be \(a - bi\). In this case,

• Root \( \alpha = \frac{1 + i\sqrt{3}}{2} \)
• Root \( \beta = \frac{1 - i\sqrt{3}}{2} \)

These roots reflect the solutions to the polynomial when the discriminant \( b^2 - 4ac \) is negative. Solving equations that involve complex roots often require understanding properties of complex numbers, like taking powers and recognizing patterns.

Powers of Complex Numbers

When dealing with powers of complex numbers, particularly for complex conjugate roots, certain patterns emerge that can simplify calculations. This is where the properties and behaviors of complex numbers and their powers become helpful.
In our problem, we observe:

• \( \alpha^0 + \beta^0 = 2 \)
• \( \alpha^1 + \beta^1 = 1 \)
• \( \alpha^2 + \beta^2 = -1 \)
• \( \alpha^3 + \beta^3 = -1 \)

We notice a repeating cycle: \(2, 1, -1, -1\). This cyclical behavior allows us to predict values for large powers by determining their position within this repeated cycle.
For example, to find \( \alpha^{2009} + \beta^{2009} \), we determine \(2009 \mod 3\), which equals 2. Thus, the result matches \( \alpha^2 + \beta^2 \), giving us -1 as the solution. Understanding these cycles can significantly reduce computational complexity when handling large exponents of complex numbers.