Data Sufficiency

[Math Revolution GMAT math practice question]

Does x^2+px+q = 0 have a root?

1) p<0
2) q<0

Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Does x^2+px+q = 0 have a root?

1) p<0
2) q<0

\[?\,\,\,:\,\,\,\,\Delta = \left( {{\text{}}{b^2} - 4ac{\text{}}} \right) = {p^2} - 4q\,\,\,\mathop \geqslant \limits^? \,\,\,\,0\,\,\]
\[\left( 1 \right)\,\,\,\left\{ \begin{gathered}
\,Take\,\,\left( {p,q} \right) = \left( { - 1,0} \right)\,\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \,\, \hfill \\
\,Take\,\,\left( {p,q} \right) = \left( { - 1,1} \right)\,\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\]
\[\left( 2 \right)\,\,\,q < 0\,\,\,\,\mathop \Rightarrow \limits^{{\text{FOCUS!}}} \,\,\, - 4q > 0\,\,\,\,\mathop \Rightarrow \limits^{{{\text{p}}^{\text{2}}}\,\, \geqslant \,\,0} \,\,\,{p^2} - 4q > 0\,\,\,\, \Rightarrow \,\,\,\left\langle {{\text{YES}}} \right\rangle \]

Conclusion: the correct answer is (B).

The above follows the notations and rationale taught in the GMATH method.

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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 discriminant of the equation is p^2 - 4q. If the discriminant is greater than or equal to zero, then the quadratic equation has roots.
The question asks if p^2-4q ≥ 0 or not.

Since p^2 ≥ 0, if q < 0, then p^2-4q ≥ 0. Thus, condition 2) is sufficient.

Condition 1)
If p = -1 and q = 0, then the discriminant is positive and the equation has 2 roots, which are 0 and 1. So, the answer is 'yes'.
If p = -1 and q = 1, then the discriminant is negative and the equation has no real roots. So, the answer is 'no'.
Since we don't have a unique solution, condition 1) is not sufficient.

Therefore, B is the answer.
Answer: B