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Source: — Data Sufficiency |
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by uwhusky » Mon Aug 16, 2010 11:51 pm
I'll give it a shot, but I am almost certain I am doing this the slow way.
In a xy plane, at what two points 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)
y = x^2 + bx + ax + ab. x-intersect means that y = 0. So the goal here is to solve 0 = x^2 + bx + ax + ab.

1) x^2 + x(a + b) + ab = x^2 + (-x) + ab; 0 = x^2 + (-x) + ab; Insufficient, we need to know ab.

2) when x = 0, y = -6; (0)^2 + b(0) + a(0) + ab = -6; ab = -6. Insufficient, we don't know what b or a is.

Together sufficient, because we just needed to know what ab is to solve 1), and 2) provides exactly that.

0 = x^2 + (-x) + ab ---> 0 = x^2 -x -6 ---> (x + 2)(x - 3), x = -2, 3.
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by kvcpk » Tue Aug 17, 2010 2:15 am
jsasipriya wrote:In a xy plane, at what two points 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)

OA C

Can someone help me with the above qn?
y=(x+a)(x+b)
intersects the x-axis when y=0
Hence x= -a or x=-b
Hence the points are (0,-a), (0,-b)

1) a+b = -1
doesnt tell what a,b are.
INSUFF

2) the graph intersects y-axis at (0, -6)
whn x=0, y=ab
Hence -6=ab
again no info about individual values of a,b.
INSUFF

Combining:
a+b=-1
and ab=-6
2 eqtns in 2 variables.
We will get a,b values.

SUFF

pick C

hope this helps!!
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don't be afraid of failure and don't abandon it.
People who work sincerely are the happiest."
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