Problem 154 - A Complete Resource Book in Mathematics for JEE Main 2017

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

2

Step 1: Identify the nature of the roots

The equation given is a quadratic equation: \(x^2 - x + 1 = 0\). The roots of a quadratic equation \(ax^2 + bx + c = 0\) can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = -1\), and \(c = 1\). Substitute these values to find the discriminant \(b^2 - 4ac\).

Step 2: Compute the discriminant

The discriminant \(b^2 - 4ac\) for the quadratic equation \(x^2 - x + 1 = 0\) is \((-1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3\).

Step 2: Determine the roots using the discriminant

Since the discriminant is negative, the roots are complex and conjugates. The roots are \(\alpha = \frac{1 + i\sqrt{3}}{2}\) and \(\beta = \frac{1 - i\sqrt{3}}{2}\). These are cube roots of unity, satisfying \(\alpha^3 = 1\) and \(\beta^3 = 1\).

Step 3: Use properties of cubic roots of unity

Given \(\alpha^3 = 1\) and \(\beta^3 = 1\), raise these equations to correspond to the powers relevant to the problem, \(\alpha^{2009}\) and \(\beta^{2009}\). Since \(2009 \equiv 0 \pmod{3}\), \(\alpha^{2009} = (\alpha^3)^{669} = 1^{669} = 1\) and \(\beta^{2009} = (\beta^3)^{669} = 1^{669} = 1\).

Step 4: Calculate and conclude

Finally, compute \(\alpha^{2009} + \beta^{2009}\). From the previous step, we found \(\alpha^{2009} = 1\) and \(\beta^{2009} = 1\). Therefore, \(\alpha^{2009} + \beta^{2009} = 1 + 1 = 2\).

Key Concepts

Complex Roots

When dealing with quadratic equations, the concept of complex roots emerges when the discriminant, given by the formula \((b^2 - 4ac)\), is negative. This indicates that the solutions to the equation are not real numbers. Instead, they take the form of complex numbers, which include an imaginary part. In simpler terms, a quadratic equation produces complex roots when you cannot find real solutions.Let's take a closer look at complex numbers. A complex number is expressed as \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit. The imaginary unit is defined by the useful property \(i^2 = -1\). For example, in the quadratic equation \(x^2 - x + 1 = 0\) from the problem, the discriminant calculated is \(-3\). This negative value confirms that the roots are complex.

• Complex roots are always conjugates of each other when originated from a quadratic equation. Conjugate means if one root is \(a + bi\), the other will be \(a - bi\).
• The roots derived in the original exercise, \(\alpha = \frac{1 + i\sqrt{3}}{2}\) and \(\beta = \frac{1 - i\sqrt{3}}{2}\), are indeed complex conjugates.

Cube Roots of Unity

In mathematics, cube roots of unity refer to the complex numbers that satisfy the equation \(x^3 = 1\). These roots are pivotal in solutions involving powers, as they help simplify calculations involving repeated multiplication of these roots.There are three cube roots of unity:

• 1, the trivial or real cube root.
• \(\omega = \frac{-1 + i\sqrt{3}}{2},\) a non-trivial complex cube root.
• \(\omega^2 = \frac{-1 - i\sqrt{3}}{2},\) another non-trivial complex cube root.

A fascinating property of these roots is that \(\omega^3 = 1\) and \(\omega^2 + \omega + 1 = 0\). For the roots found in the exercise, they are also cube roots of unity:

• \(\alpha = \omega\),
• \(\beta = \omega^2\).

Knowing this, when powers are raised significantly, such as \(\alpha^{2009}\) or \(\beta^{2009}\), they simplify based on this cyclical pattern. Because \(2009 = 3 \times 669\), it follows that these high powers reduce to 1, \(\alpha^{2009} = 1\) and \(\beta^{2009} = 1\). This reduction simplifies complex calculations immensely.

Quadratic Formula

The quadratic formula is a vital tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). It provides a direct way to compute the roots of such equations, negating the need for factoring or graphing. The standard quadratic formula is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]The formula reveals critical aspects about the roots' nature through the discriminant, \(b^2 - 4ac\):

• If positive, the equation has two distinct real roots.
• If zero, there is exactly one real root (a double root).
• If negative, as in our exercise, the roots are complex.

Exploring the quadratic \(x^2 - x + 1 = 0\), substitute coefficients \(a = 1\), \(b = -1\), and \(c = 1\) into the formula. Solving reveals the discriminant \(-3\), confirming the presence of complex roots. From this, the roots are expressed using the quadratic formula, leading to solutions that include the imaginary unit \(()i\), a hallmark of complex numbers.The quadratic formula not only assists in finding these roots, but it also prepares students to anticipate the nature of solutions before calculating this detailed answer.