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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]

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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)
[Reveal] Spoiler: OA

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]

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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
[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.

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]

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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]

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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

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Re: GMAT Prep Question, Any Shortcuts? [#permalink]

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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
[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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03 May 2011, 06:50
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03 May 2011, 10:19
y = x^2 + (a+b)x + ab is a parabola.Intersecting x axis at two points.
hence we need to find the roots of this equation.

a. a+b = -1 means a= -1 | b = -1 | a=b= -0.5 each etc. Hence not sufficient.

b y = -6 for x = 0

means ab = -6 (substituting in the original quadratic equation.

a= -3,b=1 | a = -6, b = 1 etc. hence not sufficient.

together 1+2

y = x^2 -x -6 = (x-3) * (x + 2) hence two points for X axis.

thus C.
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03 May 2011, 10:21
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y=(x+a)(x+b) = x^2+(a+b)x+ab
this is parabola equation. intersects x-axis means y=0

st 1--> a+b = -1
x^2-x+ab=0 cannot solve the equation; not sufficient

st 2--> graph intersects y-axis at -6
i.e. ab = -6
x^2+(a+b)-6=0 cannot solve the eq, not sufficient.

Both together, x^2-x-6=0
Solving gives x = -2 or x=3

Hence C.
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03 May 2011, 18:33
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udaymathapati wrote:
Attachment:
M-Q19.JPG

At what two points does the graph of y = (x+a)(x+b) intersect the x axis?

You don't need to worry what the equation represents. Just think, what does 'intersection with x axis' imply? It means the y co-ordinate is 0.

0 = (x+a)(x+b)
or x = -a or -b
Hence the graph must intersect the x axis at points (-a, 0) and (-b, 0). We need the values of a and b now.

Statement 1: a + b = -1
Two variables, only one equation. Not sufficient.

Statement 2: Graph intersects the y axis at (0, -6).
At y axis, x = 0. This means when x = 0, y co-ordinate is -6.
Put these values in y = (x+a)(x+b) to get -6 = ab.
Again, two variables, one equation. Not sufficient alone.

Using both statements, we have two variables and two different equations so we will be able to find the values of a and b. It doesn't matter which is 'a' and which is 'b'. We find that the two of them are -3 and 2. Since we need the points (-a, 0) and (-b, 0), the required points are (3, 0) and (-2, 0). Sufficient.

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In xy plane, at what 2 points does the graph of y=(x+a)(x+b) [#permalink]

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24 Jul 2011, 13:06
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In 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)

[Reveal] Spoiler:

Last edited by fluke on 24 Jul 2011, 16:37, edited 1 time in total.
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24 Jul 2011, 16:37
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tt2011 wrote:
In 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)

1) y=(x+a)(x+b)

Question is asking for the solution of x
(x+a)(x+b)=0
x^2+ax+bx+ab=0
x^2+(a+b)x+ab=0

1) a+b=-1
We got a+b. Yet, we don't know about "ab"
Not Sufficient.

2) Intersect y at (0,-6)
x=0; y=-6
-6=(0+a)(0+b)
ab=-6
We got ab. We don't know a+b.

Together;
Sufficient.

Ans: "C"
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24 Jul 2011, 21:22
We need to find the values of x when x = a or when x =b

Stmt 1: a+b = -1 gives numerous values of a and b . Insuff

Stmt 2: plugging in (0, -6) in the equation:

ab= -6 still cannot determine values of x

Combining 1 and 2 gives values of x as a quadratic eqn forms:

x^2-x-6=0. Hence C
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25 Jul 2011, 00:33
thanks fluke...i arrived at x=-a or x= -b knowing we need to find the value of x..then tried plugging in numbers..your approach is better...
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26 Oct 2011, 03:04
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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, 07:17, edited 2 times in total.
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Re: Intersection at X axis? [#permalink]

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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.

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Re: Intersection at X axis? [#permalink]

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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]

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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]

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27 Feb 2012, 00:03
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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]

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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.

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30 Apr 2012, 12:17
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.

Here's another practice question on quadratics for practice.
http://gmat.magoosh.com/questions/120
When 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: Question   [#permalink] 30 Apr 2012, 12:17

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