Bunuel wrote:

Is x^2 - 4/3*x + 5/12 < 0 ?

(1) 0 <= x

(2) x is an integer

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:This is a very tricky fraction with quadratics. We know that the lead coefficient, the coefficient of x-squared, is positive, so this is an “upward-facing” parabola, curved side pointing down and “arms” going up.

We can compute that at x = 0, y = +5/12, a positive number. At x = 1, we get y = 1 - 4/3 + 5/12 = 1/12.

At both x = 0 and x = 1, the output is positive. If a = 1 is the coefficient of x-squared, and ( – 4/3) is the coefficient of x, then the line of symmetry of the parabola is given by x = -b/(2a) = (4/3)/2 = 2/3.

The vertex, the lowest point on the parabola, would be at that position. The y-value there would be: y = (2/3)^2 - 4/3*2/3 + 5/12 = 5/12 - 4/3.

We don’t need to continue the calculation any further. A number less than one minus a fraction greater than one will be negative. At x = 2/3, there is a vertex with negative y-value, but the parabola curves up, and by the time it gets to x = 0 or x = 1, it’s already positive and above the x-axis. OK, now we can look at the statements.

Statement #1: 0 ≤ x

Well, if x = 1, then we get a “no” answer, but if x = 2/3, then we get a “yes” answer. Two different answers to the prompt are possible. This statement, alone and by itself, is not sufficient.

Statement #2: x is an integer

This is interesting and subtle. We need to think about the shape of a parabola. This parabola is negative at x = 2/3 and the immediate vicinity, but by the time we get over to either adjacent integer, x = 0 or x = 1, the parabola is already positive.

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Once a parabola is going up, it keeps going, so for any integer to the left of x = 0, or any integer to the right of x = 1, it also will be positive. It is positive for every integer value. That means, we can give a definitive answer of “no” to the prompt question. Because we were able to give a definitive answer, this statement, alone and by itself, is sufficient.

Answer = (B)

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