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JEE Main 2021
Quadratic Equations
Quadratic Equation and Inequalities
Medium

Question

If α,β,γ,δ\alpha, \beta, \gamma, \delta are the roots of the equation x4+x3+x2+x+1=0x^{4}+x^{3}+x^{2}+x+1=0, then α2021+β2021+γ2021+δ2021\alpha^{2021}+\beta^{2021}+\gamma^{2021}+\delta^{2021} is equal to :

Options

Solution

Key Concepts and Formulas

  • Roots of Unity: The nn-th roots of unity are the nn complex solutions to the equation zn=1z^n = 1.
  • Sum of Powers of Roots of Unity: If ω0,ω1,,ωn1\omega_0, \omega_1, \dots, \omega_{n-1} are the nn-th roots of unity, then k=0n1ωkp={nif p0(modn)0if p≢0(modn)\sum_{k=0}^{n-1} \omega_k^p = \begin{cases} n & \text{if } p \equiv 0 \pmod{n} \\ 0 & \text{if } p \not\equiv 0 \pmod{n} \end{cases}
  • Factorization: xn1=(x1)(xn1+xn2++x+1)x^n - 1 = (x-1)(x^{n-1} + x^{n-2} + \dots + x + 1)

Step-by-Step Solution

Step 1: Identifying the Roots

The given equation is x4+x3+x2+x+1=0x^4 + x^3 + x^2 + x + 1 = 0. We aim to recognize these roots as related to roots of unity.

Consider the equation x51=0x^5 - 1 = 0. We can factor this as: x51=(x1)(x4+x3+x2+x+1)x^5 - 1 = (x-1)(x^4 + x^3 + x^2 + x + 1)

The roots of x51=0x^5 - 1 = 0 are the 5th roots of unity, given by ei2πk5e^{i\frac{2\pi k}{5}} for k=0,1,2,3,4k = 0, 1, 2, 3, 4. These are 1,ei2π5,ei4π5,ei6π5,ei8π51, e^{i\frac{2\pi}{5}}, e^{i\frac{4\pi}{5}}, e^{i\frac{6\pi}{5}}, e^{i\frac{8\pi}{5}}.

The roots of x4+x3+x2+x+1=0x^4 + x^3 + x^2 + x + 1 = 0 are the 5th roots of unity excluding 1. This is because when x=1x=1, x4+x3+x2+x+1=50x^4 + x^3 + x^2 + x + 1 = 5 \neq 0.

Therefore, α,β,γ,δ\alpha, \beta, \gamma, \delta are the non-unity 5th roots of unity.

Step 2: Applying the Sum of Powers Property

We want to find α2021+β2021+γ2021+δ2021\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021}.

Let's consider the sum of the 2021st powers of all 5th roots of unity (including 1): 12021+α2021+β2021+γ2021+δ20211^{2021} + \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021}

Here, n=5n = 5 (number of roots of unity) and p=2021p = 2021 (the power). We need to check if pp is a multiple of nn.

Divide 20212021 by 55: 2021=5404+12021 = 5 \cdot 404 + 1 So, 20211(mod5)2021 \equiv 1 \pmod{5}. Since the remainder is 1, 20212021 is not a multiple of 55.

Applying the sum of powers of roots of unity property: 12021+α2021+β2021+γ2021+δ2021=01^{2021} + \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = 0

Step 3: Solving for the Required Sum

Since 12021=11^{2021} = 1, we have: 1+α2021+β2021+γ2021+δ2021=01 + \alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = 0

Subtracting 1 from both sides gives: α2021+β2021+γ2021+δ2021=1\alpha^{2021} + \beta^{2021} + \gamma^{2021} + \delta^{2021} = -1

Common Mistakes & Tips

  • Recognizing roots of unity: Learn to quickly recognize the polynomial xn1+xn2++x+1x^{n-1} + x^{n-2} + \dots + x + 1 as being intimately linked to the nn-th roots of unity.
  • Remember to exclude 1: Ensure you correctly identify which roots belong to the original equation. In this case, the root x=1x=1 must be excluded.
  • Modular Arithmetic: Use modular arithmetic to quickly determine if the power is a multiple of nn.

Summary

By recognizing the roots of the given equation as the non-unity 5th roots of unity, we can leverage the property of the sum of powers of roots of unity. Since 20211(mod5)2021 \equiv 1 \pmod{5}, the sum of the 2021st powers of all 5th roots of unity is 0. This allows us to isolate the required sum, resulting in an answer of -1.

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

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