Multiple choice

f(x) = ax2 + bx + c & a < 0. The equation f(x) = 0 has two distinct roots which is from the set { −2, −1, 0, 1, 2}. How many different pairs of roots of f(x) are possible such that f(0) is greater than or equals to 0?.

  1. 4

  2. 6

  3. 8

  4. 10
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C Correct answer
Explanation

For f(x) = ax^2 + bx + c with a < 0, the parabola opens downward. Roots are from {-2, -1, 0, 1, 2}. f(0) = c >= 0. Since the roots are distinct, we test pairs (r1, r2) such that f(x) = a(x-r1)(x-r2). f(0) = a(-r1)(-r2) = a(r1*r2). Since a < 0, we need r1*r2 <= 0. Possible pairs (r1, r2) from the set are (-2,-1), (-2,0), (-2,1), (-2,2), (-1,0), (-1,1), (-1,2), (0,1), (0,2), (1,2). Pairs with product <= 0 are (-2,0), (-2,1), (-2,2), (-1,0), (-1,1), (-1,2), (0,1), (0,2). Total 8 pairs.

AI explanation

Given a is less than 0, the product of the roots c must be greater than or equal to 0 for f(0) to be greater than or equal to 0. We evaluate distinct pairs of roots from the set (-2, -1, 0, 1, 2). Pairs with a non-negative product include (-2, -1), (-2, 0), (-1, 0), (0, 1), (0, 2), (1, 2), (-1, 2), and (-2, 1). Counting these valid pairs gives a total of 8 different possibilities.

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