At how many points the graph y = ax^2+bx+c intersect x-axis?

Question

At how many points the graph y = ax^2+bx+c intersect x-axis? (1) a < 0 (2) c = 6 Source: https://www.GMATinsight.com

yashikaaggarwal · Answer

1) A is negative
D = √b^2-2ac
D = √B^2+2ac (since A is less than 0)
But B and C can be Positive, negative or Zero
Giving three values; 
When D is positive = 2 real roots (will intersect X axis at 2 points)
When D is negative = no real roots (will intersect X axis at 2 points)
When D is 0 = 1 real root
Which will intersect X axis at 1 point.
(Not sufficient)

2) C is positive
D = √b^2-2ac
But B and A can be Positive, negative or Zero
Giving three values; 
When D is positive = 2 real roots (will intersect X axis at 2 points)
When D is negative = no real roots (will intersect X axis at 2 points)
When D is 0 = 1 real root
Which will intersect X axis at 1 point.
(Not sufficient)

Combining both
A is negative and C is positive.
D = √b^2-2ac
D = √B^2+2ac
Irrespective of B being positive, negative or zero.
Determinant will be positive 
And Positive determinant give two real roots. Which intersect X axis at 2 points.
(Sufficient)

Answer is C

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TestPrepUnlimited · Answer

We have a general expression for a parabola. A parabola must intersect the x-axis at 0, 1, or 2 points.

Statement 1:

With a < 0, we know the parabola opens downwards however we need to judge where the vertex is. Insufficient.

Statement 2:

c = 6 means the parabola must intersect (0, 6). However, this doesn't tell us how many points intersect x-axis, we may still have 0, 1, or 2.

Combined:

We have a parabola opening downwards that intersects (0, 6). If we want a parabola opening downwards to intersect the x-axis at 1 point only, that parabola cannot have any points higher than the x-axis so it's not possible to have this parabola intersect the x-axis at only 1 point. 0 points is also not possible as the parabola opens downwards. Therefore it must intersect the x-axis at 2 points. Sufficient.