Multiple choice

Let $a=\cos { \cfrac { 2\pi }{ 7 } } +i\sin { \cfrac { 2\pi }{ 7 } } ,\alpha =a+{ a}^{ 2 }+{ a}^{ 4 }$ and $\beta ={ a }^{ 3 }+{ a }^{ 5 }+{ a }^{ 6 } $. Then, the equation whose roots are $\alpha ,\beta $ is

  1. $\quad { x }^{ 2 }-x+2=0$
  2. ${ x }^{ 2 }+x-2=0$
  3. ${ x }^{ 2 }-x-2=0$
  4. ${ x }^{ 2 }+x+2=0$
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D Correct answer
Explanation

Given a = exp(i*2pi/7). alpha and beta are sums of powers of a. Using properties of roots of unity, alpha + beta = -1 and alpha * beta = 2. The quadratic equation is x^2 - (sum)x + (product) = 0, which is x^2 + x + 2 = 0.

AI explanation

Since a is a 7th root of unity (excluding 1), the sum of all roots is 0, meaning 1 + a + a^2 + a^3 + a^4 + a^5 + a^6 = 0. Therefore, the sum alpha + beta = (a + a^2 + a^4) + (a^3 + a^5 + a^6) equals -1. To find the product alpha * beta, we expand the terms to get the sum of all distinct pairings of powers, which equals a^3 + a^5 + a^6 + a^4 + a^6 + a^7 + a^5 + a^7 + a^9 + a^7 + a^8 + a^11, and since a^7 = 1, this simplifies to 3 + 2(a + a^2 + a^3 + a^4 + a^5 + a^6). Substituting the sum of powers as -1, the product becomes 3 + 2(-1) = 2. Forming the quadratic equation x^2 - (sum)x + (product) = 0, we get x^2 + x + 2 = 0.

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