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Source: — Data Sufficiency |

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by Rahul@gurome » Sun Oct 10, 2010 5:11 pm
Is it "at what point" or "at what two points"?
When the graph intersects the x-axis, the intersection point will be (x, 0). This implies (x + a)(x + b) = 0 or x^2 + (a + b)x + ab = 0.

(1) a + b = -1 does not gives us the values of a and b.
So, (1) is NOT SUFFICIENT.

(2) When the graph intersects the y-axis, the intersection point is (0, y). So, -6 = (0 + a)(0 + b) implies ab = -6. Again we don't get a unique answer.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), a + b = -1 and ab = -6 implies x^2 - x - 6 = 0 or (x - 3)(x + 2) = 0 or x -2, 3
So, the graph intersects at (-2, 0) and (3, 0). Assuming that it is two points, [spoiler]the correct answer is (C)[/spoiler].
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by lokesh r » Mon Oct 11, 2010 11:15 am
Rahul@gurome wrote:Is it "at what point" or "at what two points"?
When the graph intersects the x-axis, the intersection point will be (x, 0). This implies (x + a)(x + b) = 0 or x^2 + (a + b)x + ab = 0.

(1) a + b = -1 does not gives us the values of a and b.
So, (1) is NOT SUFFICIENT.

(2) When the graph intersects the y-axis, the intersection point is (0, y). So, -6 = (0 + a)(0 + b) implies ab = -6. Again we don't get a unique answer.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), a + b = -1 and ab = -6 implies x^2 - x - 6 = 0 or (x - 3)(x + 2) = 0 or x -2, 3
So, the graph intersects at (-2, 0) and (3, 0). Assuming that it is two points, [spoiler]the correct answer is (C)[/spoiler].
I got the exact solution, i read question as at what point and marked answer as E. I think it was at what points. Sorry i have not copied question correctly.
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