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by Mumbai » Mon Jun 06, 2011 3:37 am
Hi, can you guys help me with this DS question?


In the X-Y plane, at what 2 points does the graph of y = (x+a)(x+b) intersect the x-axis?
1. a+b = -1
2. The graph intersects y-axis at (0,-6)

Answer is - C - both statements together are sufficient
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Source: — Data Sufficiency |
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by galaxian » Mon Jun 06, 2011 4:07 am
Since it is a Quadratic Eqn (2 degress) in x, it will intersect the X axis at 2 points :
x = -a &
x = -b

So we just need to find the values of a & b.

From 1 - We cant find a & b............ Not Sufficient.

From 2 -
Expanding the original Eqn, we get :

(x+a)(x+b)
=> x^2 + (a+b)x +ab = 0
Putting coordinates (0,-6) in this, we get :
ab = -6 ..............................Not Sufficient.

Combining both we can find a & b.
Hence C
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by Anurag@Gurome » Mon Jun 06, 2011 4:11 am
Mumbai wrote:Hi, can you guys help me with this DS question?


In the X-Y plane, at what 2 points does the graph of y = (x+a)(x+b) intersect the x-axis?
1. a+b = -1
2. The graph intersects y-axis at (0,-6)

Answer is - C - both statements together are sufficient
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, the correct answer is [spoiler](C)[/spoiler].
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by Mumbai » Mon Jun 06, 2011 4:52 am
thanks Galaxian & Anurag, this helps :)
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by cans » Mon Jun 06, 2011 7:10 am
y=(x+a)(x+b)
intersect x-axis means y=0
thus x=-a or -b
1. a+b=-1. one equation, 2 variables. insufficient.
2. (0,-6). when x=0, ab=-6. One equation, 2 variables. Insufficient
both 1&2) a+b=-1 and ab=-6
a=-3,b=2
Points(3,0) and (-2,0) (i.e. (-a,0),(-b,0))
IMO C
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by aftableo2006 » Mon Jun 06, 2011 8:41 pm
thanks for the great explanation
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