Max@Math Revolution wrote:[GMAT math practice question]

In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts?

1) b^2-4ac<0
2) a<0

For any parabola of the form y = ax² + bx + c, the DISCRIMINANT is equal to b² - 4ac.
If the discriminant is positive, then the parabola has two x-intercepts.
If the discriminant is equal to 0, then the parabola has one x-intercept.
If the discriminant is negative, then the parabola has no x-intercepts.

Statement 1:
Since the discriminant is negative, the parabola has no x-intercepts.
Thus, the answer to the question stem is NO.
SUFFICIENT.

Statement 2:
No way to determine whether the discriminant is positive, negative, or equal to 0.
INSUFFICIENT.

The correct answer is A.

=>
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

For y=ax^2+bx+c to have an x-intercept, we must have b^2-4ac ≥ 0.
Thus, condition 1) gives the answer of 'no'. Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)
If a = -1, b = 0, c = 0, we have y = -x, which has an x-intercept of zero, so the answer is 'yes'.
If a = -1, b = 0, c = -1, we have y = -x^2 - 1, which has no x-intercept. The answer is 'no' in this case.
Since we don't have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A