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In the xy-plane, at what two points does the graph of [#permalink]
 21 Jan 2007, 00:36
Question Stats:
72% (01:47) correct
27% (00:44) wrong based on 77 sessions
In the xy-plane, at what two points does the graph of y= (x+a)(x+b) intersect the x-axis? (1) a + b = -1 (2) The graph intersects the y-axis at (0, -6) OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-xy-plane-at-what-points-does-the-graph-of-y-88398.html
Last edited by Bunuel on 01 Jun 2013, 03:28, edited 2 times in total.
Edited the question and added the OA
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For me (C) 
y= (x+a)(x+b) intersect the x-axis is equivalent to search the solutions of :
(x+a)(x+b) =0
<=> x^2 + (a+b)*x + a*b = 0 (1)
From 1
a+b = -1
implies:
(1) <=> x^2 -x + a*(-a-1) = 0
we need to know a.
INSUFF.
From 2
(0, -6) is the Y interceptor. That implies:
a*b = -6.
we need another equation with a & b, in other words, the values of a and b.
INSUFF.
Both (1) and (2)
a+b = -1
a*b = -6
These equations give us (a,b), what we need to find the roots.
SUFF.
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Re: DS: Coordinate Geometry (intersection) [#permalink]
19 Aug 2008, 08:58
lordw wrote: In the xy plane at what two points does the graph of y= (x+a) (x+b) intersect the x axis?
(1) a+b= -1
(2) The graph intersects the y axis at (0,6).
Guys I felt lost in this problem with all the letters. Any idea of how to put this in a more "visible" way? For instance, may be picking numbers? Many tks. Lw. Graph intersects x-axis when y=0 , therefore two points where graph intersects is x=-a, x=-b Statement 1) is insuff Statement 2) just tells one point Combining both , points are 6 and -5 Answer is C)
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Re: DS: Coordinate Geometry (intersection) [#permalink]
19 Aug 2008, 10:00
In the xy plane at what two points does the graph of y= (x+a) (x+b) intersect the x axis?
(1) a+b= -1
(2) The graph intersects the y axis at (0,6).
for point intersect the x axis, we need to find x when y = 0
so 0 = (x+a) (x+b)
x = -a or -b
So, we need to find a and b
(1) +(2) is sufficient to find two unknown
(C)
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The answer is (E) y=(x+a)(x+b) y=x^2+ax+ab+ab y=x^2+x(a+b)+ab statement 1 a+b = -1 y=x^2-x+ab 0=x^2-x+ab insufficient statement 2 y=x^2+x(a+b)+ab ab=-6 0=x^2+x(a+b)-6 insufficient both statements 0=x^2-x-6 6=x(x-1) x can be -2 or 3 insufficient
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I agree with fresinha12, I get 3 and -2
in the question, it asks for points, so I guess it is okay to say that the answer is C
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redbeanaddict wrote: In the xy plane, at what points does the graph of y=(x+a)(x+b) intersect the x axis?
1. a+b = -1
2. The graph intersects the y axis at (0, -6) the graph intersects x axis hen y=0 => x=-a and x=-b are the points to get the points we must know a and b but how to find 1) does not help 2)does not help => gives ab=-6 combine 1 and 2 we get a and b hence is SUFFI to answr IMO C
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Re: GMAT PREP_DS_XY plane [#permalink]
20 Aug 2009, 19:06
The answer provided above is incorrect.
The question does not ask for the values of a and b.
It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.
Statement 2 is sufficient, as it says that ab=-6.
Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.
Answer is B.
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graph question - gmat prep [#permalink]
22 Oct 2009, 13:14
Hi, so here's what I did: y= x^2+ xb+xa+ab I replace y with -6 and x with 0 -6=0+0+ab so ab= -6 I replace ab with -6 and y with 0 in the first equation 0= x^2+ xb+xa+(-6) 6=x(x+b+a) from statement A, we know that a+b=1 so 6 = x(x+1) x^2+x-6 = 0 (x-3)(x+2) x=3 or x = -2 I then got stuck  so what is the value of x?
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Re: graph question - gmat prep [#permalink]
22 Oct 2009, 13:28
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In the 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) X-intercepts of the function f(x) or in our case the function (graph) y=(x+a)(x+b) is the value(s) of x for y=0. So basically the question asks to find the roots of quadratic equation (x+a)(x+b)=0. (x+a)(x+b)=0 --> x^2+bx+ax+ab=0 --> x^2+(a+b)x+ab=0. Statement (1) gives the value of a+b, but we don't know the value of ab to solve the equation. Statement (2) tells us the point of y-intercept, or the value of y when x=0 --> y=(x+a)(x+b)=(0+a)(0+b)=ab=-6. We know the value of ab but we don't know the value of a+b to solve the equation. Together we know the values of both a+b and ab, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph. Answer: C. Hope it's clear.
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Last edited by Bunuel on 22 Oct 2009, 13:47, edited 1 time in total.
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Re: graph question - gmat prep [#permalink]
22 Oct 2009, 13:45
Bunuel wrote: In the 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)
Q is in what two points graph intersect the x-axis. Which means we should be able to find two points: (x1,0) and (x2,0), so basiacally the roots of equation: (x+a)(x+b)=0.
y=(x+a)(x+b)=x^2+x(a+b)+ab. --> We must now the values of (a+b) and ab.
(1) a+b=-1 --> not sufficient.
(2) x=0, y=-6 --> -6=ab not sufficient
(1)+(2) a+b=-1, ab=-6 --> y=x^2+x(a+b)+ab=x^2-x-6 --> x intersection means y=0 --> 0=x^2-x-6 --> (x+2)(x-3)=0 -
-> x1=-2, x2=3. Sufficient
Answer C. OMG!! I didnt realize that I need to find two points!! thats why when i ended up with -2 and 3, i got confused because i was left with two values!!! how did you find ab=-6? (2) x=0, y=-6 --> -6=ab not sufficientdid you use the formula y=mx+b? thanks,
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Re: graph question - gmat prep [#permalink]
22 Oct 2009, 13:51
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Re: graph question - gmat prep [#permalink]
23 Oct 2009, 04:34
Great Bunuel! I have been seeing your post and I have to say I am quite amazed at your math skills! Have you taken the GMAT yet? I would very much like to know your score.
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Re: graph question - gmat prep [#permalink]
23 Oct 2009, 10:47
Yup Bunuel is awesome as always Here, we need to find the solution for (x+a)(x+b) = 0. Solutions are x=-a, x=-b. Thus we need to find unique values for a and b to answer the question. 1) we get a relation between a and b but not sufficient to derive unique values of a and b. insuff. 2) we get a point on the graph, so substituting values of x and y in the graph we get ab = -6. Again, this is insufficient to find unique values of a and b. insuff. Combining, we have, a+b=-1 and ab= -6, two variables, two equations. So answer is C. The points will be (-a,0), (-b,0).
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Re: GMAT PREP_DS_XY plane [#permalink]
23 Nov 2009, 19:31
powerka wrote: The answer provided above is incorrect.
The question does not ask for the values of a and b.
It asks for the values of x when y=0. In other words, it asks for the roots of the quadratic.
Statement 2 is sufficient, as it says that ab=-6.
Thus: x^2-6x-6=0 -> Roots are -0.87 and 6.87.
Answer is B. Fig is correct. You made a mistake on the equation, it is x^2-(a+b)x+ab. Essentially, you are looking for a and b itself, or equations that tell you a+b and ab. Hence, C
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Re: graph question - gmat prep [#permalink]
25 Dec 2010, 04:49
Bunuel wrote: In the 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)
X-intercepts of the function f(x) or in our case the function (graph) y=(x+a)(x+b) is the value(s) of x for y=0. So basically the question asks to find the roots of quadratic equation (x+a)(x+b)=0.
(x+a)(x+b)=0 --> x^2+bx+ax+ab=0 --> x^2+(a+b)x+ab=0.
Statement (1) gives the value of a+b, but we don't know the value of ab to solve the equation.
Statement (2) tells us the point of y-intercept, or the value of y when x=0 --> y=(x+a)(x+b)=(0+a)(0+b)=ab=-6. We know the value of ab but we don't know the value of a+b to solve the equation.
Together we know the values of both a+b and ab, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.
Answer: C.
Hope it's clear. I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer.Can you please help on this.
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Re: graph question - gmat prep [#permalink]
25 Dec 2010, 08:49
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gautamsubrahmanyam wrote: Bunuel wrote: In the 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)
X-intercepts of the function f(x) or in our case the function (graph) y=(x+a)(x+b) is the value(s) of x for y=0. So basically the question asks to find the roots of quadratic equation (x+a)(x+b)=0.
(x+a)(x+b)=0 --> x^2+bx+ax+ab=0 --> x^2+(a+b)x+ab=0.
Statement (1) gives the value of a+b, but we don't know the value of ab to solve the equation.
Statement (2) tells us the point of y-intercept, or the value of y when x=0 --> y=(x+a)(x+b)=(0+a)(0+b)=ab=-6. We know the value of ab but we don't know the value of a+b to solve the equation.
Together we know the values of both a+b and ab, hence we can solve the quadratic equation, which will be the x-intercepts of the given graph.
Answer: C.
Hope it's clear. I get your explanation.But using a+b=-1 and ab=-6 and solving it I get (a-b)^2=25 .which means a-b=+5 or -5.So since I cant be sure of the value of a-b,I selected option e as the answer. Can you please help on this. The question asks: at what two points the graph of y=(x+a)(x+b) intersect the x-axis? So we should find two points: (x1, 0) and (x2, 0), (points of intersection of the given graph with X-axis). Basically the question asks to find the roots of quadratic equation (x+a)(x+b)=0 --> x^2+(a+b)x+ab=0. When we combine statement we have: a+b=-1 and ab=-6, so you should solve quadratic equation x^2-x-6=0 --> x_1=-2 and x_2=3 --> points of intersection are (-2, 0) and (3, 0). Check on the diagram: Attachment: 
MSP272919df1bed26600c1g0000673adafe5fce1fc2.gif [ 3.61 KiB | Viewed 7540 times ]
Hope it's clear.
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This problem is easy when you don't start by rapidly trying to solve it.
Think about it before you put pen to paper...
"what two points does the graph intersect the x axis..."
This is a simple problem asking for 2 values of x...
statement 1 gives us a+b=-1
statement 2 gives us ab
Start with the stem...multiply x+a and x+b, it will become very clear.
Y = X^2 + X (a+b) + ab
statement 1 gives you a+b, we need ab
statement 2 says graph intersects y at (0,-6). Plug this into the original statement, you will get ab.
Solved = C!
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