Question

If $\alpha, \beta, \gamma, \delta$ are the roots of the equation $x^4 - 4x^3 + 3x^2 + 2x - 2 = 0$ such that $\alpha$ and $\beta$ are integers and $\gamma, \delta$ are irrational numbers, then $\alpha + 2\beta + \gamma^2 + \delta^2 =$

• A. 5
• B. 7
• C. 11
• D. 13

Hint:

When dealing with higher-degree polynomials, always start by checking for simple integer or rational roots using the Rational Root Theorem. Once a root is found, use synthetic division to reduce the degree of the polynomial. This makes it easier to find the remaining roots. For expressions involving sums of powers of roots like $\gamma^2+\delta^2$, it's usually faster to use Vieta's formulas than to calculate the roots explicitly.

Answer 1

The Correct Option is C

Step 1: Find Integer Roots Possible integer roots are divisors of the constant term -2: \( \pm 1, \pm 2 \). - \( P(1) = 1 - 4 + 3 + 2 - 2 = 0 \). Root found: 1. - \( P(-1) = 1 + 4 + 3 - 2 - 2 \neq 0 \). - \( P(2) = 16 - 32 + 12 + 4 - 2 \neq 0 \). Perform synthetic division by \( (x-1) \): Resulting cubic: \( x^3 - 3x^2 + 2 = 0 \). Check divisors of 2 for this cubic: - \( Q(1) = 1 - 3 + 2 = 0 \). Root found: 1. So, \( x=1 \) is a root with multiplicity at least 2. The integer roots are \( \alpha=1, \beta=1 \).
Step 2: Find Irrational Roots Divide \( x^3 - 3x^2 + 2 \) by \( x-1 \): Resulting quadratic: \( x^2 - 2x - 2 = 0 \). Roots are \( \gamma, \delta = \frac{2 \pm \sqrt{4+8}}{2} = 1 \pm \sqrt{3} \) (Irrational).
Step 3: Calculate Expression We need \( \alpha + 2\beta + \gamma^2 + \delta^2 \). \( \alpha = 1, \beta = 1 \). From \( x^2 - 2x - 2 = 0 \), \( \gamma + \delta = 2 \) and \( \gamma\delta = -2 \). \( \gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma\delta = (2)^2 - 2(-2) = 4 + 4 = 8 \). Total Sum = \( 1 + 2(1) + 8 = 11 \).