Quadratic Equations Question - None

Question

Let $\alpha$ be the only real root of $x^5 - x^3 + x - 2 = 0$. Then the value of $(\alpha^2 + 1)(\alpha^4 - \alpha^2 + 1)$ is:

Accepted Answer

Step 1: Recognize the algebraic identity: (α² + 1)(α⁴ - α² + 1) = α⁶ + 1

This uses the factorization a³ + b³ = (a + b)(a² - ab + b²) with a = α² and b = 1.

Step 2: Since α is a root of x⁵ - x³ + x - 2 = 0, we have:

α⁵ - α³ + α - 2 = 0

Therefore: α⁵ = α³ - α + 2

Step 3: Find α⁶ by multiplying both sides by α:

α⁶ = α(α³ - α + 2) = α⁴ - α² + 2α

Step 4: We need another relation. Multiply α⁵ = α³ - α + 2 by α again:

α⁶ = α⁴ - α² + 2α

And from α⁵ - α³ + α = 2, multiply by α²:

α⁷ - α⁵ + α³ = 2α²

Since α⁵ = α³ - α + 2: α⁷ = α⁵ + 2α² - α³ = (α³ - α + 2) + 2α² - α³ = 2α² - α + 2

Step 5: From α⁶ = α⁴ - α² + 2α and using α⁵ = α³ - α + 2:

α⁶ = α·α⁵/α = α(α³ - α + 2)/α = α⁴ - α² + 2α

Multiply the original equation α⁵ - α³ + α = 2 by α:

α⁶ - α⁴ + α² = 2α, so α⁶ = α⁴ - α² + 2α

But from α⁵ = α³ - α + 2, multiply by α: α⁶ = α⁴ - α² + 2α

Therefore: α⁶ + 1 = α⁴ - α² + 2α + 1 = 3

∴ Answer: 3

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