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# Gmat perp DS-2

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Manager
Joined: 20 Mar 2005
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Gmat perp DS-2 [#permalink]  07 Nov 2007, 05:00
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
please explain this ..i dont have the OA.
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Manager
Joined: 25 Jul 2007
Posts: 108
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Kudos [?]: 15 [0], given: 0

When the given graph will intersect the x-axis, the value of y will be 0.

Therefore, we have

(x+a)(x+b)=0

which means that either x=-a or x=-b will be the x-co-ordinate of the point of intersection.

We thus need to find the value of a & b. Statements 1 & 2 are two equations with 2 variables. We can thus get the values of a & b.
The two points of intersection will be (-a,0) & (-b,0).

Important note: we get two pairs of values for a & b. However, the value of the x co-ordinate doesn't change in this case. The points are (3,0) and (-2,0).
Manager
Joined: 01 Nov 2007
Posts: 69
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Kudos [?]: 3 [0], given: 0

I get two values of a and two values of b

so total four points of intersection of the graph with X-axis
Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA
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Kudos [?]: 15 [0], given: 0

y=(x+a) (x+b)
y=(x^2 + ax + bx + ab)
y=(x^2 +x (a+b) + ab

from 1 (a+b)= -1 insuff
from 2 -6=0+0+ab insuff

Togather

at point of intersection y=0
0=x^2 + (-1)x + (-6) = x= (-1 + sqrt (1+24))/2 or (-1 - sqrt (1+24))/2
SVP
Joined: 28 Dec 2005
Posts: 1579
Followers: 2

Kudos [?]: 91 [0], given: 2

I end up with E....

at the end, all i know is the two values are some -3 and +2 .... but we dont know which one a is, and which one b is
Manager
Joined: 25 Jul 2007
Posts: 108
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Kudos [?]: 15 [0], given: 0

I don't think anyone read the Important Note I wrote in my post.

I will explain again.

Now we know that for the points of intersection at the x-axis , y=0.
Therefore,from the given equation in the question,
we have (x+a)(x+b)=0.

You do get 2 values for 'a' & 'b'. However, the co-ordinates of the points of intersection do not change. You get exactly two possible co-ordinates.

The two possible values of 'a' are -3 & 2.
The two possible values of 'b' are 2 & -3.

i.e either a= -3 & b=2
OR
a=2 & b=-3.

Consider scenario 1.
if a=-3 & b=2, we have
(x-3)(x+2)=0. (since y=0)
Therefore x=3 OR x=-2.
Therefore the points of intersection would be (3,0) and (-2,0)

Consider scenario 2.
if a=2 & b=-3,we have
(x+2)(x-3)=0
Therefore x=-2 OR x=3
Therefore the points of intersection would again be (3,0) and (-2,0)

Please note that the actual values of 'a' & 'b' are irrelevant. The question asks for the 2 points of intersection and we can derive them from the given information.
CEO
Joined: 29 Mar 2007
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Re: Gmat perp DS-2 [#permalink]  07 Nov 2007, 22:10
Balvinder wrote:
please explain this ..i dont have the OA.

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

We are looking for point (x,0)

1: a+b=-1. No help here.

2: Y int: (0,-6) No help here either b/c

-6=(x+a)(x+b) plug in for x ---> ab=-6. We cannot find out the x int. w/ this.

1&2: y=x^2+xa+xb+ab -- > x(x+a+b) -6 ---> x(x-1)-6 --> x^2-x-6

(x-3)(x+2) the line crosses the x axis at either point 3 or -2. Insuff.

E
SVP
Joined: 28 Dec 2005
Posts: 1579
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Kudos [?]: 91 [0], given: 2

you know what, after reading it again this morning, i change my answer to C.

Jbs is right in his post above; using both statements we can come up with values for a and b, but really you are ending up with just two points, (-3,0) and (2,0)
Director
Joined: 03 Sep 2006
Posts: 884
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Kudos [?]: 288 [0], given: 33

y = (x+a)(x+b) = x^2 + (a+b)x +ab

This is the equation of right side up Parabola. ( Positive general quadratic equation )

y = ax^2 + bx + c ( general positive quadratic equation )

y = a(x – h)^2 + k, then the vertex is the point (h, k). This makes sense, if you think about it. The squared part is always positive (for a right-side-up parabola).

y = ax^2 + bx + c, the vertex (h, k) is found by computing h = –b/2a, and then evaluating y at h to find k.
k = (4ac – b^2) / 4a

Just the background if someone wants to know fundamentals

Okkkkk so the equation we have is

y = x^2 + (a+b)x +ab

vertex (h,k) can be found

h= -(a+b)/2

k=(4ab-(a+b)^2)/4

from (i) we get the vertex as ( -1/2, ab-1/4)

As we do not know ab, we can't determine the point where it corsses the X axis.

From (ii), we know the value of ab=-6

Combining (i) and (ii), we can definitely find the points where the parabola crosses the x-axis. ( I am not finding, but the points would be possibly ( not necessary ) coresponding points on the positive side and negative side of the X axis )

*************************************************************

Second simple and layman's approach

equation y=x^2+(a+b)x+ab

from (i) y= x^2-x+ab

For y=0, x= ab and also x = 1-ab

as we don't know value of ab, we can't say what are the two values of x

From(ii), we get ab=-6

From (i) and (ii), we can definitely find the values of x where the parabola instersects.

x=-6 and x= 7
Senior Manager
Joined: 27 Aug 2007
Posts: 253
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Kudos [?]: 10 [0], given: 0

Another vote for C,
we know to points, either (-3;2) and (2;-3) one of them

Ans: C
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