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sud21: Master | Next Rank: 500 Posts; Posts: 344; Joined: Sat Nov 12, 2011 3:21 amFollowed by: 2 members; Thanked: 1 times

by sud21 » Thu Jan 19, 2012 11:03 pm

In xy-plane, Y=ax^2+bx+c,does the graph intersect with X axis?
1). a>0
2). c<0

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Source: Beat The GMAT — Quantitative Reasoning | Thu Jan 19, 2012 11:03 pm

avik.ch: Legendary Member; Posts: 641; Joined: Wed Apr 20, 2011 1:15 amFollowed by: 32 members; Thanked: 149 times; GMAT Score:760

by avik.ch » Fri Jan 20, 2012 2:25 am

Are you sure this is a GMAT problem. This problem need the concept of determinent.

d = b^2 - 4 ac

if b^2 - 4 ac > 0, then it intersects x axis twice.
if b^2 - 4 ac = 0, then x = -b/2a, so it intersects x axis once.
if b^2 - 4 ac <0 , it will not touch x axis. So no solution.

By now way we can check the second condition as we dont know the value of a,b and c.

1. a>0

b^2 - 4ac can be greater or less than 0, depending on the value of c.insufficient

2. c < 0

b^2 - 4ac can be greater or less than 0, depending on the value of a.insufficient

combinig both :

b^2 = positive, a>0 and c<0

so the equation is b^2 - 4a(-c) = b^2 + 4ac , so b^2 + 4ac > 0 ( both a and c are positive)

hence sufficient.

IMO:C

hope this helps !!

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Jim@StratusPrep: MBA Admissions Consultant; Posts: 2279; Joined: Fri Nov 11, 2011 7:51 amFollowed by: 266 members; Location: New York; Thanked: 660 times; GMAT Score:770

by Jim@StratusPrep » Fri Jan 20, 2012 12:05 pm

A touch easier way to think of this:

'a' will determine whether or not the parabola opens upward or downward.

'b' is meaningless to this problem because it shifts the parabola left and right

'c' This determines the vertex of the parabola --> the value of the 'y-coordinate'

(1) If a is negative then it is a parabola that faces downward. --> NS
(2) Determines vertex --> NS

Together they are sufficient because the parabola opens upward with a negative vertex, meaning it crosses the x-axis twice.

Keep these problems simple!

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