ziyuenlau wrote:How many intersects with x-axis does y=x^2+2qx+r have ?

(1) q^2 > r

(2) r^2 > q

OFFICIAL ANSWER : A

Hi ziyuenlau,

Since y=x^2+2qx+r intersects with x-axis, y=0.

Thus, x^2+2qx+r = 0

x^2+2qx+r = 0 is a quadratic equation whose discriminant is (2q)^2 - 4r = 4q^2 -4r.

You must know that for a quadratic equation ax^2 + bx + c = 0, the discriminant = b^2 - 4ac.

1. If the discriminant: b^2 - 4ac = 0, we have only one solution.

2. If the discriminant: b^2 - 4ac > 0, we have two solutions.

3. If the discriminant: b^2 - 4ac < 0, we have no real solution.

We have to see if 4q^2 -4r < = > 0 or q^2 -r < = > 0 or q^2 < = > r

Statement 1: q^2 > r

It is clear that q^2 > r, thus, there are two intersects with x-axis. Sufficient.

Statement 2: r^2 > q

Since r can be negative or positive, we cannot decide. Insufficient.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Geometry Guide

-Jay
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