Question

One of the roots of the equation \( x^{14} + x^9 - x^5 - 1 = 0 \) is:

• A. \(\frac{1 + \sqrt{3}i}{2}\)
• B. \(\frac{\sqrt{5} - 1}{4} + i\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)
• C. \(\frac{1 - \sqrt{3}i}{2}\)
• D. \(\frac{\sqrt{5} + 1}{4} + i\frac{\sqrt{10 - 2\sqrt{5}}}{4}\)

Hint:

For polynomial equations with complex roots, consider roots of unity and trigonometric identities for simplification.

Answer 1

The Correct Option is D

To solve the equation \( x^{14} + x^9 - x^5 - 1 = 0 \), we can factorize it and analyze its roots. Step 1: Factorize the equation The given equation is: \[ x^{14} + x^9 - x^5 - 1 = 0. \] We can factorize it as follows: \[ x^{14} + x^9 - x^5 - 1 = x^9(x^5 + 1) - 1(x^5 + 1) = (x^9 - 1)(x^5 + 1). \] Thus, the equation becomes: \[ (x^9 - 1)(x^5 + 1) = 0. \] This gives two cases: 1. \( x^9 - 1 = 0 \), or 2. \( x^5 + 1 = 0 \). Step 2: Solve \( x^9 - 1 = 0 \) The equation \( x^9 - 1 = 0 \) has roots that are the 9th roots of unity: \[ x = e^{2k\pi i/9}, \quad k = 0, 1, 2, \dots, 8. \] These roots lie on the unit circle in the complex plane. Step 3: Solve \( x^5 + 1 = 0 \) The equation \( x^5 + 1 = 0 \) has roots that are the 5th roots of \(-1\): \[ x = e^{(2k+1)\pi i/5}, \quad k = 0, 1, 2, 3, 4. \] These roots also lie on the unit circle in the complex plane. Step 4: Check the given options We now check which of the given options is a root of the equation. Option (1): \( \frac{1 + \sqrt{3}i}{2} \) This is a primitive 3rd root of unity. It is not a root of \( x^9 - 1 = 0 \) or \( x^5 + 1 = 0 \), so it is not a solution. Option (2): \( \frac{\sqrt{5} - 1}{4} + i\frac{\sqrt{10 - 2\sqrt{5}}}{4} \) This is a primitive 5th root of \(-1\). It not satisfies \( x^5 + 1 = 0 \), so it is is not a solution. Option (3): \( \frac{1 - \sqrt{3}i}{2} \) This is a primitive 3rd root of unity. It is not a root of \( x^9 - 1 = 0 \) or \( x^5 + 1 = 0 \), so it is not a solution. Option (4): \( \frac{\sqrt{5} + 1}{4} + i\frac{\sqrt{10 - 2\sqrt{5}}}{4} \) This is a primitive 5th root of unity. It satisfies \( x^5 - 1 = 0 \), and \( x^5 + 1 = 0 \), so it is a solution.