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The question can be rephrased: when is x^2+2qx+r = 0

Using the quadratic formula we find that x = (-2q +/- sqrt(4q^2 - 4r)/2)

From this we can see that there will be two solutions if 4q^2 - 4r > 0, one solution if 4q^2 - 4r = 0 and zero solutions if 4q^2 - 4r < 0.

From the information in 1.: If q^2 > r then 4q^2 - 4r > 0.

SUFFICIENT - there are two points of intersection on the y axis.

From the information in 2.: If r^2>q we don't know if 4q^2 - 4r is positive. We try two cases to demonstrate this: if q = 1 and r = -2 then r^2>q and 4q^2 - 4r > 0 if q = 1 and r = 2 then r^2>q but 4q^2 - 4r < 0

This information is INSUFFICIENT.

So I chose A.

gmatclubot

Re: How many intersects with x-axis
[#permalink]
26 Nov 2011, 18:22

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