xy plane graph intersect the x-axis

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xy plane graph intersect the x-axis

by skprusty » Thu Dec 09, 2010 10:14 pm
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now sure how to approach this type of problem

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by 4GMAT_Mumbai » Fri Dec 10, 2010 2:54 am
Hi,

Thanks ... Interesting question ...

1. -a and -b are the points where the curve meets the x-axis. However, we do not know their numeric values

a + b = -1

Implies a = 5 and b = -6

or

a = 10 and b = -11 (besides the millions of other possibilities). Hence, Insuff.

2.

y = (x + a) (x + b)

If (0,-6) lies on this graph, then

-6 = (0+a) * (0+b)

ab = -6

This implies a = -3 and b = 2

or

a = -6 and b = 1 (besides other possibilities). Hence, insuff

3. Combine both

a + b = -1

ab = -6

Here, a = -3 and b = 2

or

a = 2 and b = -3

However, the curve becomes y = (x + 2) (x - 3). The curve passes through (-2,0) and (3,0). Hence, suff.

My take is C. Please correct / confirm.

Thanks !
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by Rahul@gurome » Fri Dec 10, 2010 4:34 am
Given: y = (x + a)(x + b)
Implies the graph intersects x-axis at x = -a or x = -b.
Thus we have to find the values of a and b.

Statement 1: (a + b) = -1
Infinite numbers of values are possible for a and b.

Not sufficient.

Statement 2: The graph intersects y-axis at (0, -6)
For x = 0, y = (0 + a)(0 + b) = ab = -6
Infinite numbers of values are possible for a and b.

Not sufficient.

1 and 2 Together: (a + b) = -1 and ab = -6
Now, (a - b)² = (a + b)² - 4ab = (-1)² - 4*(-6) = 25
=> (a - b) = ±5

Thus, either (a = 2, b = -3) or (a = -3, b = 2)
In both of the cases the graph intersects x-axis at x = -3 and x = 2.

Sufficient.

The correct answer is C.
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by skprusty » Fri Dec 10, 2010 9:07 am
thanks OA: C