Question

The sum of two roots of the equation \(x^4 - x^3 - 16x^2 + 4x + 48 = 0\) is zero. If \(\alpha, \beta, \gamma, \delta\) are the roots of this equation, then \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4\) is:

• A. \(123\)
• B. \(369\)
• C. \(132\)
• D. \(396\)

Hint:

Utilize symmetry in the roots, especially when given that some roots are additive inverses, to simplify calculations involving powers or sums of roots.

Answer 1

The Correct Option is B

To solve the problem, we are given the quartic equation: \[ x^4 - x^3 - 16x^2 + 4x + 48 = 0, \] and we know that the sum of two of its roots is zero. Let the roots be \(\alpha, \beta, \gamma, \delta\), with \(\alpha + \beta = 0\). We are tasked with finding the value of \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4\). Step 1: evaluate the calculation Using the identity for the sum of fourth powers of roots: \[ \alpha^4 + \beta^4 + \gamma^4 + \delta^4 = (\alpha + \beta + \gamma + \delta)^4 - 4(\alpha + \beta + \gamma + \delta)^2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta) \]\[+ 4(\alpha + \beta + \gamma + \delta)(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta) + 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)^2 - 4\alpha\beta\gamma\delta. \] Substitute the known values: \[ \alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 1^4 - 4(1)^2(-16) + 4(1)(-4) + 2(-16)^2 - 4(48). \] Simplify: \[ \alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 1 + 64 - 16 + 512 - 192 = 369. \] Final Answer: \[ \boxed{369} \]