Bunuel wrote:
For integers x, b and c, x² + bx + c = 0. Is x > 0?
(1) b > 0
(2) c > 0
Target question: Is x > 0? Statement 1: b > 0 Let's TEST some cases that satisfy statement 1:
If b = 5, and c = -14, the equation becomes x² + 5x - 14 = 0. Factor to get: (x + 7)(x - 2), which means EITHER x = -7 OR x = 2
If x = -7, then the answer to the target question is
NO, x is NOT greater than 0If x = 2, then the answer to the target question is
YES, x is greater than 0Since we cannot answer the
target question with certainty, statement 1 is NOT SUFFICIENT
Statement 2: c > 0Let's TEST some cases that satisfy statement 2:
Case a: b = -6, and c = 9. The equation becomes x² - 6x + 9 = 0. Factor to get: (x - 3)(x - 3), which means x = 3. So, the answer to the target question is
YES, x is greater than 0Case b: b = 6, and c = 9. The equation becomes x² + 6x + 9 = 0. Factor to get: (x + 3)(x + 3), which means x = -3. So, the answer to the target question is
NO, x is NOT greater than 0 Statements 1 and 2 combined We have: x² + bx + c = 0
Let's say that, when we factor x² + bx + c we get: x² + bx + c = (x + j)(x + k)
If c is POSITIVE, we know that EITHER j and k are both positive, OR j and k are both negative
We also know that, since b is POSITIVE, the sum of j and k must be POSITIVE
This allows us to rule out the possibility that j and k are both negative, since two negative numbers cannot add up to be positive.
So, it MUST be the case that j and k are both positive
In other words, x² + bx + c = (x + positive)(x + positive)
If (x + positive)(x + positive) = 0, then we can see that two solutions will both be negative.
For example, if (x + 2)(x + 3) = 0, then the two solutions are x = -2 and x = -3
Similarly, if (x + 1)(x + 10) = 0, then the two solutions are x = -1 and x = -10
If the two solutions will both be NEGATIVE, the answer to the target question is
NO, x is NOT greater than 0Since we can answer the
target question with certainty, the combined statements are SUFFICIENT
Answer: C
Cheers,
Brent
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