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In the XY, at what two points does the graph of y

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Intern
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Joined: 25 Dec 2008
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In the XY, at what two points does the graph of y [#permalink]

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New post 25 Dec 2008, 10:11
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

In the XY, at what two points does the graph of y =(X+a)(X+b) intersect the x-axis?

1) a+b=-1
2) the graph intersect y-axis at (0,-6)

pls explain this

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Re: source GMAT CAT [#permalink]

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New post 25 Dec 2008, 11:16
y = (x+a)(x+b)

statement 2
-------------

Graph intercept Y axis at (0, -6). As X = 0 on y -axis
If we put this values in the given eq.

ab = -6 ------- 1st eq

statement 1
------------

a+b = -1 => b = -(a+1) ..put this value in above equation

-(a+1)a = -6
a can be 2 OR -3,
so b can be -3 OR 2

so now if we put this 2 combinations of a & b into the given equation

y = (x + 2)(x-3) ----- 2nd eq

When this eq intercepts at X-axis than y =0

so (x+2)(x-3) = 0
x^2 -x -6 = 0 =>

x = 3 OR -2

so 2 points on x -axis is
(3, 0) & (-2, 0)

ANS is ######## (C) ###############
Hope it clears !

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New post 25 Dec 2008, 11:38
thanks GUPTA

can you make it little bit more clear

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Re: source GMAT CAT [#permalink]

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New post 25 Dec 2008, 20:11
I think you can read the problem in this way:
as the graph of function y= (x+a)(x+b) cut x-axis at two points (-a,0) and (-b,0)( except the case a=b) because y= 0 if x= -a OR x= -b.
Therefore, you have to find this two points, right. So just try to find the two equations relating to two variables a and b.

Statement 1 : clearly an equation a+b = -1

Statement 2: the graph cut y-axis at (0,-6) so when x = 0 then y = (0+a)(0+b) = ab = -6

so, simultaneous equation a+b = -1 [1]
ab= -6 [2]
From [1] we get a= -1-b, and then replace it to [2] we get:
b(-1-b) = -6 => b(1+b) = 6

Solve this equation you get, b = 2 OR b = -3
IF b = 2 then a = -3
IF b = -3 then a= 2

So in these two situations, we get the same interception points (2,0) and (-3,0)

So from two statement you find the results, so it is C.

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Re: source GMAT CAT   [#permalink] 25 Dec 2008, 20:11
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