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

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Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post 11 Sep 2018, 02:49
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A
B
C
D
E

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[Math Revolution GMAT math practice question]

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

\(1) a<0\)
\(2) c>0\)

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Re: Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post Updated on: 11 Sep 2018, 11:14
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

\(1) a<0\)
\(2) c>0\)
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



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Originally posted by sumit411 on 11 Sep 2018, 06:22.
Last edited by sumit411 on 11 Sep 2018, 11:14, edited 1 time in total.
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Re: Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post 11 Sep 2018, 08:23
I don't think the ans is c. As mentioned by the poster above, we need details of B

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Re: Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post 11 Sep 2018, 10:58
1
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

\(1) a<0\)
\(2) c>0\)


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.

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Re: Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post 11 Sep 2018, 12:03
MathRevolution wrote:
[Math Revolution GMAT math practice question]

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

\(1) a<0\)
\(2) c>0\)


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

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

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

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

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


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

Regards,
fskilnik.
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Re: Does y=ax^2+bx+c intersect the x-axis?  [#permalink]

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New post 13 Sep 2018, 01:34
=>

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

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Re: Does y=ax^2+bx+c intersect the x-axis?   [#permalink] 13 Sep 2018, 01:34
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