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In the xy plane, at what points does the graph of y= [#permalink]
 24 Dec 2009, 10:00
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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)
Last edited by Bunuel on 26 Feb 2013, 05:23, edited 2 times in total.
Edited the question and added the OA
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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
24 Dec 2009, 10:16
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JimmyWorld wrote: This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve? 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) OA Thanks. 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. For more on this topic check Coordinate Geometry chapter of Math Book (link in my signature). Hope it helps.
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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
24 Dec 2009, 10:37
y=(x+a)(x+b) when y=0 To solve this one, what do we need to know? Obviously a or b, which are not stated in the information (1) & (2) So one rule advise by MGMAT Book, always expand when the information given is factorized or Factorized when the information given is expended. We know from a quadratic expression, the x axis intersect when y=0 So let's expand, and we have x^2 + (a + b)x + ab = 0 So now, we need to know ab and (a+b) to solve this equation Therefore the correct answer is C, since only both information taken together permit to answer the question.
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Re: DS question : need help [#permalink]
15 Oct 2010, 01:05
satishreddy wrote: need help on DS question Right, so when the graph intersects the x-axis, y is 0. Those points are clearly x=-a and x=-b (1) a+b=1. Not sufficient to know a or b (2) We have point (0,-6). Plus it into the equation to get -6=ab. Again not sufficient to know either number (1+2). Now we have 2 equations, sufficient. a-6/a=1; a^2-a-6=0; a=3 or a=-2 Hence b=-2 or b=3 .. Either way the points are 3 and -2 Answer is (c)
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Re: GMAT Prep Question, Any Shortcuts? [#permalink]
02 Nov 2010, 07:47
JimmyWorld wrote: This question took me much longer then 2 minutes to figure out and sucked up a lot of time for me. Are there any shortcuts that would make this problem easier to solve? 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) OA Thanks. We know that the graph will intersect the x-axis when y=0, which occurs when x = -a and x = -b. 1) This doesn't tell us what a and b are. For example, a could be 5 and b could be -6, or a could be -4 and b could be 3. Insufficient. 2) This tells us that either a or b is 6, but we don't know the other point. Insufficient. Together, we know that one of the values is 6, and that a + b = -1. So we can get the other value. Sufficient.
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Re: Points of intersection [#permalink]
26 Oct 2011, 03:04
kotela wrote: In the XY plane 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)
can anyone plz explain? Intersect the x axis means y is 0. the x coordinates will be -a and -b respectively. 1. a+b=-1 Insufficient 2. -6=(0+a)(0+b) ab=-6 Insufficient. 1+2 a+b=-1 ab=-6 a=-3, b=2 a=2, b=-3 Sufficient as we want two points we can say (-3,0) and (-2,0) without knowing the exact values of a & b.
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Last edited by blink005 on 26 Oct 2011, 07:17, edited 2 times in total.
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Re: Intersection at X axis? [#permalink]
16 Feb 2012, 18:41
Here is my attempt:
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).
Essentially they are asking us to solve for x when y=0:
y = (x+a)(x+b) -> x^2 + xb + xa + ab = 0 (What is X?)
1) a = -b - 1 -> This is insufficient because it still leaves us with one unknown: b
2) -6 = ab -> This allows us to solve for ab in the line equation, but we still don't know what a and b are individually, which we need because we have an xb and xa term.
1) + 2) -> Knowing a = -6/b, we can plug into 1) and find b and a separately. Then the only unknown is x which we can then solve for.
Answer: C
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Re: Intersection at X axis? [#permalink]
16 Feb 2012, 18:58
The graph intersects at 0,-6. That is when x = 0, and y=-6. X being zero conveniently cancels out all the terms in our original equation:
x^2 + xb + xa + ab = y
and we are left with ab = -6
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Re: In the xy plane, at what points does the graph of y= [#permalink]
26 Feb 2012, 23:50
I tried plugging in values in this one.
Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.
Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.
Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C.
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Re: In the xy plane, at what points does the graph of y= [#permalink]
27 Feb 2012, 00:03
mourinhogmat1 wrote: I tried plugging in values in this one.
Started from option 2:
ab = -6
Possible values are (2,-3); (3,-2); (-6,1); (1,-6) --> INSUFFICIENT.
Option 1:
substitute for x and y in equation we get
a+b = -1
Several possible values such as (-3,2); (-8,7) and so on. --> INSUFFICIENT.
Combining both --> find from option 2 which possible value leads to a+b =-1, only one of the four choices does that (2,-3). Hence SUFFICIENT. Answer choice C. This is not a good idea to plug the numbers for this problem, it's better to understand the concept and you won't need any math at all (certainly you won't need to solve a+b=-1 and ab=-6). Next, there are infinitely many values of a and b possible to satisfy ab=-6, not just four: notice that we are not told that a and b are integers only, so for example a=1/2 and b=-12 is also a solution.
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Re: In the xy plane, at what points does the graph of y= [#permalink]
26 Apr 2012, 21:33
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y= x^2+ax+bx+ab
y= x^2+x(a+b)+ab
the question asks us to find the value of two points where the graph intersects x axis. So y=0
x^2+x(a+b)+ab=0 =?
1) a+b=-1 , but we dot have any value of ab, hence not sufficient
2) putting the value x=0 , y=-6
y= x^2+x(a+b)+ab
ab=-6
but we dot have any value of a+b , hence not sufficient
1&2) x^2+x(a+b)+ab=0
now we have a+b=-1
ab =-6
hence sufficient to find 2 values of x.
So C is the answer
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Stiv 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) So, it's important to know --- when the quadratic is given in factored form --- y= (x + a) (x + b) --- then we know the two roots, x = -a, and x = -b. Roots are the x-intercepts, the places where the graph intersects the x-axis. Basically, the prompt is asking us to find the values of a & b. Statement #1: a + b = -1 One equation for two unknowns. Not enough to solve. Not sufficient. Statement #2: The graph intersects the y-axis at (0, -6) Plugging in x = 0 (the condition of the y-axis), we get y = (0+a)(0+b) = ab = -6 Again, one equation for two unknowns. Not enough to solve. Not sufficient. Combined statements: a + b = -1 ab = -6 Two equations with two unknowns ---> we can solve for the values of a & b, which will answer the question. Sufficient. Answer = CHere's another practice question on quadratics for practice. http://gmat.magoosh.com/questions/120When you submit your answer to that question, the next page will have a full video explanation. Let me know if you have any further questions. Mike 
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Re: Coordinate DS; y = (x+a)(x+b) [#permalink]
02 Sep 2012, 11:50
Intersecting the x-axis meens that we have to find the roots of the equation (when will y=0?).
y = (x+a) * (x+b) = x^2 + a*x + b*x + a*b = x^2 + x*(a+b) + a*b
Statement (1) is not sufficient. We don't know specific values of a and b, nor can we substitute all variables (a and b).
Statement (2) just gives us one point.
Combining both statements:
0 = (-6)^2 + (-6)*(-1) + a*b
0 = 36 + 6 + a*b
-42 = a*b
Substituting a*b (and (a+b)):
x^2 + x*(-1) + (-42) = 0
You don't have to calculate any further. It's easy to see now that we have sufficient information to calculate the other point.
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Re: In the xy plane, at what points does the graph of y= [#permalink]
04 Sep 2012, 01:38
1 statement is clearly insufficient, since we have nothing about x or y 2 statement tells us that the ab=-6, but nothing about a or b individually Combining statements we have: a=-6/b, -6/b+b=-1, b^2+b-6=0, a=-3, b=2, so both statements together are sufficient (C)
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Coordinate geometry [#permalink]
23 Apr 2013, 01:21
In the xy 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 the y axis at (0, -6 )
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Re: Coordinate geometry [#permalink]
23 Apr 2013, 01:44
kabilank87 wrote: In the xy 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 the y axis at (0, -6 ) The function is a parabola y=x^2+x(a+b)+ab1)Does y=x^2-x+ab intersec x? It depend on ab.Not sufficient 2) Tells us that point (0,6) is on the parabol, so ab=-6. Does y=x^2+x(a+b)-6 intersect x? It depends on a+b. Not sufficient 1+2) The equation of the parabola is complete y=x^2-x-6, we can tell if it intersect x. Sufficient C
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Re: Coordinate geometry [#permalink]
23 Apr 2013, 03:27
Zarrolou wrote: kabilank87 wrote: In the xy 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 the y axis at (0, -6 ) The function is a parabola y=x^2+x(a+b)+ab1)Does y=x^2-x+ab intersec x? It depend on ab.Not sufficient 2) Tells us that point (0,6) is on the parabol, so ab=-6. Does y=x^2+x(a+b)-6 intersect x? It depends on a+b. Not sufficient 1+2) The equation of the parabola is complete y=x^2-x-6, we can tell if it intersect x. Sufficient C Hi Zarro , In equation no 2 - How you use ab=-6. Is n't ab = +6. Will you please explain the logic behind it ? Thanks in advance.
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Last edited by kabilank87 on 23 Apr 2013, 03:31, edited 1 time in total.
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Re: Coordinate geometry [#permalink]
23 Apr 2013, 03:31
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kabilank87 wrote: Hi Zarro ,
In equation no 2 - How you replace ab with -6. Will you please explain the logic behind it ?
Thanks in advance. Sure! Lets take another look at (2)The graph intersects the y axis at (0, -6 ) We know that the parabola has equation y=x^2+x(a+b)+ab, now if it intersect y in (0,-6) this point is in the parabola, so we just substitute its value (x=0,y=-6) in the equation -6=0^2+0(a+b)+ab and we find that ab=-6Hope it's clear now, let me know
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Re: Coordinate geometry [#permalink]
23 Apr 2013, 03:39
Zarrolou wrote: kabilank87 wrote: Hi Zarro ,
In equation no 2 - How you replace ab with -6. Will you please explain the logic behind it ?
Thanks in advance. Sure! Lets take another look at (2)The graph intersects the y axis at (0, -6 ) We know that the parabola has equation y=x^2+x(a+b)+ab, now if it intersect y in (0,-6) this point is in the parabola, so we just substitute its value (x=0,y=-6) in the equation -6=0^2+0(a+b)+ab and we find that ab=-6Hope it's clear now, let me know Hi, Very clear. Thank you.
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Re: In the xy plane, at what points does the graph of y= [#permalink]
24 Apr 2013, 18:10
This question is a bit tricky , if you know the following it helps :
Any quadratic equation represents a parabola in the x-y domain.
here , the equation is y=(x+a)(x+b) . So the roots of the equation are -a,-b . So it cuts the x axis at -a and -b.
1) a+b = 1 -> Doesn't help since we have one equation with 2 variables.
2) Gives you one set of value for x and y. So you can get an equation in a and b.
Combining 1 and 2 gives the answer to A and B. C wins.
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Re: In the xy plane, at what points does the graph of y=
[#permalink]
24 Apr 2013, 18:10
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