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In the xy plane, at what points does the graph of y=

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In the xy plane, at what points does the graph of y= [#permalink] New post 24 Dec 2009, 09:00
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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)
[Reveal] Spoiler: OA

Last edited by Bunuel on 26 Feb 2013, 04:23, edited 2 times in total.
Edited the question and added the OA
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Re: GMAT Prep Question, Any Shortcuts? [#permalink] New post 24 Dec 2009, 09:16
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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
[Reveal] Spoiler:
C


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] New post 24 Dec 2009, 09:37
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] New post 15 Oct 2010, 00:05
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] New post 02 Nov 2010, 06:47
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
[Reveal] Spoiler:
C


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] New post 26 Oct 2011, 02:04
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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, 06:17, edited 2 times in total.
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Re: Intersection at X axis? [#permalink] New post 16 Feb 2012, 17:41
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] New post 16 Feb 2012, 17:58
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] New post 26 Feb 2012, 22:50
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] New post 26 Feb 2012, 23:03
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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] New post 26 Apr 2012, 20:33
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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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Re: Question [#permalink] New post 30 Apr 2012, 11:17
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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 = C

Here's another practice question on quadratics for practice.
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Let me know if you have any further questions.

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Re: Coordinate DS; y = (x+a)(x+b) [#permalink] New post 02 Sep 2012, 10:50
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] New post 04 Sep 2012, 00:38
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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Re: Coordinate geometry [#permalink] New post 23 Apr 2013, 00:44
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)+ab

1)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] New post 23 Apr 2013, 02:27
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)+ab

1)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, 02:31, edited 1 time in total.
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Re: Coordinate geometry [#permalink] New post 23 Apr 2013, 02:31
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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=-6

Hope it's clear now, let me know
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Re: In the xy plane, at what points does the graph of y= [#permalink] New post 24 Apr 2013, 17:10
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] New post 25 May 2014, 03:10
I tried to get the equation by applying the special product of (x+y)^2. Is my reasoning correct that I cant do that because a is not equal to b so we basically dont have a y?!

Thanks in advance!
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Re: In the xy plane, at what points does the graph of y= [#permalink] New post 25 May 2014, 04:39
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steilbergauf wrote:
I tried to get the equation by applying the special product of (x+y)^2. Is my reasoning correct that I cant do that because a is not equal to b so we basically dont have a y?!

Thanks in advance!


Your question is not clear at all. Can you please elaborate? Thank you.
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Re: In the xy plane, at what points does the graph of y=   [#permalink] 25 May 2014, 04:39
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