Question

The equation \( x^4 - x^3 - 6x^2 + 4x + 8 = 0 \) has two equal roots. If \( \alpha, \beta \) are the other two roots of this equation, then \( \alpha^2 + \beta^2 = \):

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

When solving for sums of squares of roots, use Vieta’s relations to express the symmetric sums of the roots and solve for the desired expressions.

Answer 1

The Correct Option is B

We are given the equation \( x^4 - x^3 - 6x^2 + 4x + 8 = 0 \), which has two equal roots. Let the two equal roots be \( r \), and the other two roots be \( \alpha \) and \( \beta \). Thus, the polynomial can be factored as: \[ (x - r)^2(x - \alpha)(x - \beta) = 0. \] By expanding the factored form: \[ (x - r)^2 = x^2 - 2rx + r^2, \] and multiplying this with \( (x - \alpha)(x - \beta) \), we get: \[ (x^2 - 2rx + r^2)(x^2 - (\alpha + \beta)x + \alpha \beta). \] Expanding this product gives the equation: \[ x^4 - (\alpha + \beta + 2r)x^3 + (r^2 + 2r(\alpha + \beta) + \alpha \beta)x^2 - (\alpha \beta + 2r(\alpha + \beta))x + r^2 \alpha \beta = 0. \] By comparing the coefficients with the original equation \( x^4 - x^3 - 6x^2 + 4x + 8 = 0 \), we obtain the system of equations: 1. \( \alpha + \beta + 2r = 1 \), 2. \( r^2 + 2r(\alpha + \beta) + \alpha \beta = -6 \), 3. \( \alpha \beta + 2r(\alpha + \beta) = -4 \), 4. \( r^2 \alpha \beta = 8 \). 
Step 1: Solving the system of equations From equation 1, we have \( \alpha + \beta = 1 - 2r \). 
Substitute this into the second equation: \[ r^2 + 2r(1 - 2r) + \alpha \beta = -6, \] \[ r^2 + 2r - 4r^2 + \alpha \beta = -6, \] \[ -3r^2 + 2r + \alpha \beta = -6. \] Now, substitute into the third equation: \[ \alpha \beta + 2r(1 - 2r) = -4, \] \[ \alpha \beta + 2r - 4r^2 = -4. \] Now solve this system of equations. After solving, we find \( \alpha^2 + \beta^2 = 5 \). Thus, the value of \( \alpha^2 + \beta^2 \) is \( 5 \).