If α is an imaginary root of x5 – 1 = 0 then the equation whose roots are α + α4 and α2 + α3 is

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x2 – x – 1 = 0

x2 + x – 1 = 0

x2 – x + 1 = 0

x2 + x + 1 = 0

answer is B.

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Detailed Solution

given that α is an imaginary root of the equation x5 – 1 = 0. That is, α satisfies the equation x5 – 1 = 0. Therefore, we can write α5 − 1 = 0⋯⋯ (1)
We know that the factorization of xn − 1 is given by xn – 1 = (x −1)(1 + x + x2 + ... + xn – 2 + xn − 1). We will use this information in equation (1). Therefore, we get
α5 – 1 = 0
⇒ (α − 1) (1 + α + α2 + α3 + α4) = 0
⇒ α – 1 = 0 or 1 + α + α2 + α3 + α4 = 0 [∵ ab = 0 ⇒ a = 0 or b = 0]
⇒ α = 1 or 1 + α + α2 + α3 + α4 = 0
It is given that α is an imaginary root. So, we can say that α ≠ 1 because α = 1 is the real root. Therefore, we have only 1 + α + α2 + α3 + α4 = 0 ⋯⋯ (2). Now from the equation (2), we can write
α + α2 + α3 + α4 = −1
⇒ (α+α4) + (α2+α3) = −1⋯⋯ (3)
From equation (3), we can say that the sum of two roots (α + α4) and (α2 + α3) is −1.
Let us multiply by α on both sides of the equation (1). Therefore, we get α6 = α and α7 = α2.
Let us find the product of two roots (α + α4) and (α2 + α3). Therefore, we get
(α + α4)(α2 + α3)
= α3 + α4 + α6 + α7
= α3 + α4 + α + α2 [∵ α6 = α, α7 = α2]
= α + α2 + α3 + α4
= −1
Note that here we write α + α2 + α3 + α4 = −1 from equation (2).
Now we have the following information:
(1) The sum of two roots (α + α4) and (α2 + α3) is −1.
(2) The product of two roots (α + α4) and (α2 + α3) is −1.
If the sum and product of two roots is known then we can say that the quadratic equation will be
x2 − (sum of roots) x + (product of roots) = 0
Therefore, the required equation is
x2 − (−1)x + (−1) = 0
⇒ x2 + x – 1 = 0
Therefore, if α is an imaginary root of x5 – 1 = 0 then the equation is x2 + x – 1 = 0 whose roots are (α + α4) and (α2 + α3).
Hence, option (2) is the correct option.

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