Question

If the sum of two roots of the equation \( x^4 - 2x^3 + x^2 + 4x - 6 = 0 \) is zero then the sum of the squares of the other two roots is

• A. -6
• B. 1
• C. -2
• D. 0

Hint:

Using the symmetry of roots (like \( \alpha, -\alpha \)) often simplifies the standard Vieta's relations significantly, causing many terms to cancel out. Focus on the odd-sum relations (sum of roots, sum of product of three) to isolate variables quickly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We are given a quartic equation (degree 4). Let the roots be \( \alpha, \beta, \gamma, \delta \). The condition states that the sum of two roots is zero, say \( \alpha + \beta = 0 \), which implies \( \beta = -\alpha \).
Step 2: Key Formula or Approach:

For \( Ax^4 + Bx^3 + Cx^2 + Dx + E = 0 \): 1. \( \sum \alpha = -B/A \) 2. \( \sum \alpha\beta = C/A \) 3. \( \sum \alpha\beta\gamma = -D/A \)
Step 3: Detailed Explanation:

Let the roots be \( \alpha, -\alpha, \gamma, \delta \). 1. Sum of roots: \[ \alpha + (-\alpha) + \gamma + \delta = -\frac{-2}{1} \] \[ \gamma + \delta = 2 \] 2. Sum of roots taken three at a time (\( \sum \alpha\beta\gamma \)): The terms are \( \alpha(-\alpha)\gamma + \alpha(-\alpha)\delta + \alpha\gamma\delta + (-\alpha)\gamma\delta \). \[ -\alpha^2\gamma - \alpha^2\delta + \gamma\delta(\alpha - \alpha) = -\frac{4}{1} \] \[ -\alpha^2(\gamma + \delta) = -4 \] Substitute \( \gamma + \delta = 2 \): \[ -\alpha^2(2) = -4 \implies \alpha^2 = 2 \] 3. Sum of roots taken two at a time (\( \sum \alpha\beta \)): Terms: \( \alpha(-\alpha) + \alpha\gamma + \alpha\delta - \alpha\gamma - \alpha\delta + \gamma\delta \). Notice terms like \( \alpha\gamma \) cancel with \( -\alpha\gamma \). \[ -\alpha^2 + \gamma\delta = \frac{1}{1} = 1 \] Substitute \( \alpha^2 = 2 \): \[ -2 + \gamma\delta = 1 \implies \gamma\delta = 3 \] We need to find the sum of the squares of the other two roots, i.e., \( \gamma^2 + \delta^2 \). \[ \gamma^2 + \delta^2 = (\gamma + \delta)^2 - 2\gamma\delta \] Substitute known values: \[ \gamma^2 + \delta^2 = (2)^2 - 2(3) = 4 - 6 = -2 \]
Step 4: Final Answer:

The sum of the squares of the other two roots is -2.