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100 points for $49 worth of Veritas practice GMATs FREE VERITAS PRACTICE GMAT EXAMS Earn 10 Points Per Post Earn 10 Points Per Thanks Earn 10 Points Per Upvote ## Does y=ax^2+bx+c intersect the x-axis? tagged by: Max@Math Revolution ##### This topic has 2 expert replies and 0 member replies ### GMAT/MBA Expert ## Does y=ax^2+bx+c intersect the x-axis? ## Timer 00:00 ## Your Answer A B C D E ## Global Stats Difficult [Math Revolution GMAT math practice question] Does y=ax^2+bx+c intersect the x-axis? 1) a<0 2) c>0 _________________ Math Revolution Finish GMAT Quant Section with 10 minutes to spare. The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy. Only$149 for 3 month Online Course
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### GMAT/MBA Expert

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

Does y=ax^2+bx+c intersect the x-axis?

1) a<0
2) c>0
$?\,\,\,\,:\,\,\,\,a{x^2} + bx + c = 0\,\,\,\,{\text{has}}\,\,\left( {{\text{real}}} \right)\,\,{\text{roots?}}$
$\left( 1 \right)\,\,\,a < 0\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( { - 1,0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = - {x^2}\,\,\,{\text{parabola}}} \right] \hfill \\ \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( { - 1,0, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{No}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = - {x^2} - 1\,\,\,{\text{parabola}}} \right] \hfill \\ \end{gathered} \right.$
$\left( 2 \right)\,\,\,c > 0\,\,\,\left\{ \begin{gathered} \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {0,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = x + 1\,\,{\text{line}}} \right] \hfill \\ \,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {1,0,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{No}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = {x^2} + 1\,\,\,{\text{parabola}}} \right] \hfill \\ \end{gathered} \right.$
$\left( {1 + 2} \right)\,\,\,\,a \ne 0\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,\,:\,\,\,\,\Delta = {b^2} - 4ac\,\,\,\mathop \geqslant \limits^? \,\,\,0$
$\left. \begin{gathered} a < 0\,\, \hfill \\ c > 0 \hfill \\ \end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,\, - 4ac > 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{b^2}\,\, \geqslant \,\,0} \,\,\,\,\Delta > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUF}}.$

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

Regards,
fskilnik.

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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.

Since we are asked about a quadratic polynomial, the question asks if the discriminant b^2-4ac is positive. Neither condition 1) nor condition 2) on its own is sufficient to determine this.

Considering both conditions 1) & 2) together yields b^2-4ac > 0 since b^2 ≥ 0 and ac < 0.

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