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

This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 157
Joined: Sat Nov 19, 2016 5:34 am
Thanked: 2 times
Followed by:4 members
How many intersects with x-axis does y=x^2+2qx+r have ?

(1) q^2 > r

(2) r^2 > q

OFFICIAL ANSWER : A

Master | Next Rank: 500 Posts
Posts: 157
Joined: Sat Nov 19, 2016 5:34 am
Thanked: 2 times
Followed by:4 members

by hazelnut01 » Mon Apr 10, 2017 5:38 am
Could someone help me with this question?

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3008
Joined: Mon Aug 22, 2016 6:19 am
Location: Grand Central / New York
Thanked: 470 times
Followed by:34 members

by Jay@ManhattanReview » Mon Apr 10, 2017 6:33 am
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
_________________
Manhattan Review GMAT Prep

Locations: Almaty | Minsk | Aarhus | Vilnius | and many more...

Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.