If discriment of the quadratic equation is 0--> one solution---> touches x axis

If discernment of quadratic equation is +ve---> 2 solution---> touches x axis

If discernment of quadratic equation is - ve---> no solution---> does not touches x axis

Discrimant value = b^2- 4ac

St 1 and St2 don't give information about a and c independently

Combining: we still don't know value of b
But Discrimant is always positive


C



Thank you = Kudos

I don't think the ans is c. As mentioned by the poster above, we need details of B

Posted from my mobile device

If Discriminant D=b^2- 4ac is positive or equal to zero, then the curve y intersects x axis.

Statement 1: a<0 .No information about c. Not sufficient.

Statement 2: c>0. No information about a. Not sufficient.

Combining 1 and 2

ac is negative, so D=b^2+4ac. b^2 can be zero or positive, so either way D>0. Hence the curve does intersect the x axis.

Hence C.

Cheers!

\(?\,\,\,\,:\,\,\,\,a{x^2} + bx + c = 0\,\,\,\,{\text{has}}\,\,\left( {{\text{real}}} \right)\,\,{\text{roots?}}\)

\(\left( 1 \right)\,\,\,a < 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( { - 1,0,0} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = - {x^2}\,\,\,{\text{parabola}}} \right] \hfill \\
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( { - 1,0, - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{No}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = - {x^2} - 1\,\,\,{\text{parabola}}} \right] \hfill \\
\end{gathered} \right.\)

\(\left( 2 \right)\,\,\,c > 0\,\,\,\left\{ \begin{gathered}
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {0,1,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{Yes}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = x + 1\,\,{\text{line}}} \right] \hfill \\
\,{\text{Take}}\,\,\left( {a,b,c} \right) = \left( {1,0,1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{No}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {y = {x^2} + 1\,\,\,{\text{parabola}}} \right] \hfill \\
\end{gathered} \right.\)

\(\left( {1 + 2} \right)\,\,\,\,a \ne 0\,\,\,\,\, \Rightarrow \,\,\,\,\,?\,\,\,\,:\,\,\,\,\Delta = {b^2} - 4ac\,\,\,\mathop \geqslant \limits^? \,\,\,0\)

\(\left. \begin{gathered}
a < 0\,\, \hfill \\
c > 0 \hfill \\
\end{gathered} \right\}\,\,\,\,\, \Rightarrow \,\,\,\, - 4ac > 0\,\,\,\,\,\,\,\mathop \Rightarrow \limits^{{b^2}\,\, \geqslant \,\,0} \,\,\,\,\Delta > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,{\text{SUFF}}.\)


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

Regards,
fskilnik.
_________________

=>

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 \(ac < 0.\)

Therefore, C is the answer.
Answer: C

_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________