I'm going to assume that the problem also requires p to be an integer.
x² + px - 24 = 0
x² + px = 24
x * (x + p) = 24
So x must be a positive or negative factor of 24. That gives us x = -24, -12, -8, -6, -4, -3, -2, -1, 1, 2, 3, 4, 6, 8, 12, and 24, or sixteen values, each of which has a corresponding integer solution for p. There's some overlap, though! If x = -24, then p = 23, but if x = 1, again p = 23.
Going further, every value of p corresponds to two values of x, so we only have eight integer values of p.
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How many integer values
Source: Beat The GMAT — Quantitative Reasoning |
Whoops, should add a note explaining that last assertion. Since x * (x + p) = 24, we also have p = (24 - x)²/x. (It's safe to divide by x since x = 0 is not a solution.)
Any integer p that's a solution to that quadratic is going to itself create a quadratic with two solutions. For instance, say p = 2. Then we'd have
2 = (24 - x)²/x, or
x² + 2x - 24 = 0
which would factor as (x + 6) * (x - 4).
Since every p value is a factor of 24, every quadratic created by a p will have two integer solutions, meaning that each p corresponds to two values of x.
Any integer p that's a solution to that quadratic is going to itself create a quadratic with two solutions. For instance, say p = 2. Then we'd have
2 = (24 - x)²/x, or
x² + 2x - 24 = 0
which would factor as (x + 6) * (x - 4).
Since every p value is a factor of 24, every quadratic created by a p will have two integer solutions, meaning that each p corresponds to two values of x.
x2 + px - 24 = 0
(x+a)(x-b) = 0
ab = 24 -> positive factor pairs (a * b) are 1*24, 2*12, 3*8, 4*6, 6*4, 8*3, 12*2, 24*1
a-b = p = -23, -10, -5, -2, 2, 5, 10, 23
So there are 8 possible values for p
(x+a)(x-b) = 0
ab = 24 -> positive factor pairs (a * b) are 1*24, 2*12, 3*8, 4*6, 6*4, 8*3, 12*2, 24*1
a-b = p = -23, -10, -5, -2, 2, 5, 10, 23
So there are 8 possible values for p
