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Does the parabola y=mx^2+4mnx+4n, where m and n are constants, interse 12 Jul 2020, 08:16
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65% (hard)

Question Stats:

51% (02:05) correct 49% (02:00) wrong based on 39 sessions

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Does the parabola \(y=mx^2+4mnx+4n\), where \(m\) and \(n\) are constants, intersect the \(x\)-axis?
1) \(mn>1\)
2) \(m>n\)
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A
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Does the parabola y=mx^2+4mnx+4n, where m and n are constants, interse 13 Jul 2020, 06:44
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TheUltimateWinner
Does the parabola \(y=mx^2+4mnx+4n\), where \(m\) and \(n\) are constants, intersect the \(x\)-axis?
1) \(mn>1\)
2) \(m>n\)
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To find x-intercepts of any line or curve in coordinate geometry, we set y = 0 and solve. So we want to know if this equation

mx^2 + 4mnx + 4n = 0

has any solutions for x. The standard way in algebra to answer a question like this is to use the "discriminant" from the quadratic formula. The quadratic formula tells us the solutions to any quadratic at all, and the "discriminant" in that formula is under a square root. If the discriminant is negative, the root is undefined, and there are no solutions. Otherwise the quadratic has one or two solutions (one when the discriminant is zero, two when the discriminant is positive). You don't need to know anything about the quadratic formula or about discriminants for the GMAT, though, so the question seems beyond the scope of the test unless there's another way to solve that I'm not immediately seeing.

Anyway, the discriminant of a general quadratic ax^2 + bx + c is b^2 - 4ac, so the discriminant of the quadratic in this question is (4mn)^2 - 4(m)(4n) = 16(mn)^2 - 16mn = 16mn(mn - 1). For our quadratic to have at least one solution, and thus for the parabola to have at least one x-intercept, this needs to be at least zero:

16mn(mn - 1) > 0
mn(mn - 1) > 0

This will be false only when mn and mn-1 have opposite signs, which only happens when mn is positive and mn - 1 is negative, so when 0 < mn < 1. So if we can be sure mn does not lie between 0 and 1, we can be sure the answer to the question is 'yes'. Statement 1 clearly ensures mn is not in that range, while Statement 2 does not, so the answer is A.

I'd add, just in case someone quickly looking at the quadratic in the question thinks it's a perfect square (that's what I thought at first glance, just because of the pattern of the numbers), it is not equal to (mx + 2n)^2, because if you expand that, you get m^2 x^2 at the beginning, and not mx^2. So there's no obvious factorization we can take advantage of, which is why I had to use the quadratic formula (not something needed on the GMAT).
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Does the parabola y=mx^2+4mnx+4n, where m and n are constants, interse 13 Jul 2020, 09:39
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Parabola (x − h)^2 = 4p(y − k)-see below pic

i need to make sure multiplication of mn should be +ve in order not to cut x axis.
Here:(sqrt(m) x+ 2sqrt(n))^2

sqrt(mn)=?

Statement1: mn +ve
sufficient

Statement2: mn is+ve or -ve ; can't say:
not sufficient

hence A
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parabola.png
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parabola.png
parabola.png [ 26.27 KiB | Viewed 1527 times ]

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