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In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts?

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In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts? [#permalink]

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New post 07 Jun 2018, 02:41
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[GMAT math practice question]

In the x-y plane, does the parabola \(y=ax^2+bx+c\) have x-intercepts?

\(1) b^2-4ac<0\)
\(2) a<0\)
Spoiler: OA

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Re: In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts? [#permalink]

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New post 07 Jun 2018, 03:10
MathRevolution wrote:
[GMAT math practice question]

In the x-y plane, does the parabola \(y=ax^2+bx+c\) have x-intercepts?

\(1) b^2-4ac<0\)
\(2) a<0\)


The value \(b^2-4ac\) is called discriminant and it has the following properties:

1) If \(b^2-4ac>0\), then the equation \(ax^2+bx+c=0\) has two roots or in other words, the parabola has two \(x\)-intercepts.
2) If \(b^2-4ac<0\), then the equation \(ax^2+bx+c=0\) has no roots or in other words, the parabola does not have \(x\)-intercept.
3) If \(b^2-4ac=0\), then the equation \(ax^2+bx+c=0\) has one single root or in other words, the parabola has one single \(x\)-intercept.

(1) The parabola does not have \(x\)-intercepts. Answer is NO. Sufficient.

(2) We don't know whether the discriminant is positive, negative or zero. Not sufficient.

Answer: A
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Re: In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts? [#permalink]

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New post 07 Jun 2018, 03:15
Statement 2 only tells us whether the parabola opens upwards or downwards.

Statement 1 on the other hand tells us that the roots of the equation are imaginary. This means that the parabola won't cut the x-axis.

Hence one can say that statement 1 is sufficient while 2 alone is not.
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Re: In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts? [#permalink]

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New post 10 Jun 2018, 18:29
=>
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.

For \(y=ax^2+bx+c\) to have an x-intercept, we must have \(b^2-4ac ≥ 0\).
Thus, condition 1) gives the answer of ‘no’. Since ‘no’ is also a unique answer by CMT (Common Mistake Type) 1, condition 1) is sufficient.

Condition 2)
If \(a = -1, b = 0, c = 0\), we have \(y = -x,\) which has an x-intercept of zero, so the answer is ‘yes’.
If \(a = -1, b = 0, c = -1\), we have \(y = -x^2 – 1\), which has no x-intercept. The answer is ‘no’ in this case.
Since we don’t have a unique solution, condition 2) is not sufficient.

Therefore, A is the answer.
Answer: A
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Re: In the x-y plane, does the parabola y=ax^2+bx+c have x-intercepts?   [#permalink] 10 Jun 2018, 18:29
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