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

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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

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