lf $\alpha, \beta$ are the real roots of the equation $ax^{2}+bx+c=0$, then the value of $(1+\alpha+\alpha^{2})(1+\beta+\beta^{2})$ is
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positive
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negative
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non-negative
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non-positive
The expression (1+alpha+alpha^2)(1+beta+beta^2) can be written as (1-alpha^3)/(1-alpha) * (1-beta^3)/(1-beta). Since alpha and beta are roots of a quadratic with real coefficients, the expression evaluates to a real number. Without specific values, the sign depends on the coefficients, but in the context of such problems, it is often positive.
Expanding the expression (1+alpha+alpha squared) times (1+beta+beta squared) gives 1 plus alpha plus beta plus alpha squared plus beta squared plus alpha squared beta plus alpha beta squared plus alpha squared beta squared. We can group these terms to show it equals (alpha beta) squared plus (alpha plus beta) times (alpha beta) plus (alpha plus beta) squared plus (alpha plus beta) plus 1. Since alpha and beta are real roots, alpha plus beta is a real number, and the quadratic t squared plus t plus 1 has a negative discriminant, making it strictly positive for any real t; the product is therefore positive.