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 \( 5 \).

Answer 2

Given: The quartic equation \[ x^4 - x^3 - 6x^2 + 4x + 8 = 0 \] has two equal roots. Let the other two roots be \( \alpha \) and \( \beta \). Find the value of \[ \alpha^2 + \beta^2. \]

Step 1: Use the fact that the equation has a repeated root Since the quartic has a repeated root, its polynomial and its derivative share a common root.
Derivative: \[ 4x^3 - 3x^2 - 12x + 4 = 0. \]

Step 2: Find the repeated root Let the repeated root be \( r \). Then \( r \) satisfies both equations: \[ r^4 - r^3 - 6r^2 + 4r + 8 = 0, \] \[ 4r^3 - 3r^2 - 12r + 4 = 0. \] Test possible rational roots using factors of constant terms. Try \( r = 2 \): - Check in original polynomial: \[ 2^4 - 2^3 - 6(2^2) + 4(2) + 8 = 16 - 8 - 24 + 8 + 8 = 0, \] - Check in derivative: \[ 4(8) - 3(4) - 12(2) + 4 = 32 - 12 - 24 + 4 = 0. \] So, \( r = 2 \) is the repeated root.

Step 3: Divide the polynomial by \( (x - 2)^2 \) to get quadratic factor Divide \[ x^4 - x^3 - 6x^2 + 4x + 8 \] by \[ (x - 2)^2 = (x - 2)^2 = x^2 - 4x + 4. \] Perform polynomial division or use synthetic division twice: Dividing once: \[ x^4 - x^3 - 6x^2 + 4x + 8 \div (x - 2) = x^3 + x^2 - 4x - 4. \] Dividing the quotient by \( (x - 2) \): \[ x^3 + x^2 - 4x - 4 \div (x - 2) = x^2 + 3x + 2. \] Thus, \[ x^4 - x^3 - 6x^2 + 4x + 8 = (x - 2)^2 (x^2 + 3x + 2). \]

Step 4: Find roots \( \alpha, \beta \) from quadratic factor Solve \[ x^2 + 3x + 2 = 0. \] Factor: \[ (x + 1)(x + 2) = 0 \implies x = -1, -2. \] So, \[ \alpha = -1, \quad \beta = -2. \]

Step 5: Calculate \( \alpha^2 + \beta^2 \) \[ \alpha^2 + \beta^2 = (-1)^2 + (-2)^2 = 1 + 4 = 5. \]

Final answer: \[ \boxed{5}. \]