The quadratic function $$ f(x) = ax^2 + bx + c $$, where $$ a

Question

The quadratic function $$ f(x) = ax^2 + bx + c $$, where $$ a < 0 $$, has two distinct roots selected from the set $$\{-2, -1, 0, 1, 2\}$$. Determine how many pairs of roots are possible such that $$ f(0) \geq 0 $$.

Accepted Answer

To solve this problem, note that $$ f(0) \geq 0 $$ implies $$ c \geq 0 $$. Since $$ a < 0 $$, the product of the roots of $$ f(x) = ax^2 + bx + c $$, given by $$ c/a $$, must be negative or zero. This means the roots must either have opposite signs or one of them must be zero.

From the set of possible roots $$\{-2, -1, 0, 1, 2\}$$, we identify all pairs where the product of the roots satisfies this condition. The valid pairs are:  
$$(0, 2), (0, 1), (1, -2), (0, -1), (0, -2), (-1, 2), (-2, 2), (-1, 1)$$.

Thus, there are **8 distinct pairs** of roots satisfying the condition $$ f(0) \geq 0 $$. Therefore, the correct answer is **Option c)**.

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