Data Sufficiency

[Math Revolution GMAT math practice question]

In the xy-plane, does the graph of y=ax^2+c intersect the x-axis?

1) a>0
2) c>0

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

In the xy-plane, does the graph of y=ax^2+c (a nonzero) intersect the x-axis?

1) a>0
2) c>0

$$\left\{ {\,\left( {x,y} \right)\,\,:\,\,\,y\,\, = a{x^2} + c\,\,\,,\,\,\,a \ne 0\,} \right\}\,\,\,\, \cap \,\,\,\left\{ {\,\left( {x,y} \right)\,\,:\,\,y = 0} \right\}\,\,\,\,\,\mathop \ne \limits^? \,\,\,\,\,\emptyset \,$$




(1) a> 0 :: INSUFFICIENT : Take figure 1 (YES) and figure 2 (NO)

(2) c > 0 :: INSUFFICIENT : Take figure 2 (NO) and figure 5 (YES)

(1+2) SUFFICIENT: NO


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

P.S.: we believe the proposer´s idea was to avoid a = 0 (so that it is a "parabola problem" only).
If we are wrong, our arguments are still valid and the correct answer is still (C). Think about that!

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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.

The question "does the graph of y=ax^2+c intersect the x-axis" is equivalent to asking "does the equation ax^2+c = 0 have a root".
Note that the statement "ax^2 + bx + c = 0 has a root" is equivalent to b^2-4ac ≥ 0.
Thus, the question asks if -4ac ≥ 0, or ac ≤ 0, since b = 0 in this problem.

When we consider both conditions together, we obtain ac > 0 and the answer is "no", since a > 0 and c > 0.
Since 'no' is also a unique answer by CMT (Common Mistake Type) 1, both conditions together are sufficient.

Note: Neither condition on its own provides enough information for us to determine whether ac ≤ 0.

Therefore, C is the answer.
Answer: C