For the polynomial $p(x)=x^4-5x^3+8x^2-4x$, find the value of ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2$, where $\alpha ,\beta ,\gamma ,\delta$ are its zeros.

Step-by-step solution

1. Identify the coefficients: for $x^4+a_1{x}^3+a_2{x}^2+a_{3}x+a_4$, the sum of roots $S_1=-a_1$.
2. Here $a_1=-5$, so $S_1=5$ (sum of the four zeros).
3. The sum of pairwise products $S_2=a_2=8$.
4. Use the identity $\sum {\alpha }_i^2=S_1^2-2S_2$.
5. Compute $S_1^2=5^2=25$ and $2S_2=2\times 8=16$.
6. Subtract: $25-16=9$.
7. Hence ${\alpha }^2+{\beta }^2+{\gamma }^2+{\delta }^2=9$.

Concept tested: Sum of squares of roots using Vieta

Published 24 August 2026 · How this solution was prepared and checked

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