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New post 12 Aug 2008, 10:32
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Re: DS - GMATPREP - coordinate geometry -2 [#permalink]

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New post 12 Aug 2008, 11:22
Can anyone please explain this .... please

Also,how can I improve my cordinate geometry .....I am taking the exam in 4 days,

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Re: DS - GMATPREP - coordinate geometry -2 [#permalink]

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New post 12 Aug 2008, 11:25
mba9now wrote:
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Boss... please type the question..


intersect x axis when y=0
(x+a)(x+b)=0
--> two points are .. (-a,0) and (-b,0)

1) a+b=-1
--> one equation two variable..s

Not suffciient

2) graph intersect y at (0,-6)

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

-6=(0+a) (0+b) --> ab=-6

Not suffcieint

combine both

a+b=-1
ab=-6
a-6/a=-1

a^2-6+a=0
--> a*a+3a-2a-6 =0 --> a(a+3)-2(a+3) ===>
(a+3)(a-2)=0

a=-3 or a=2
substitue values in a+b=-1
a=-3 b=2
a=2 b=-3
(both solutions are same )

suffcieint

C
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Last edited by x2suresh on 12 Aug 2008, 11:40, edited 1 time in total.

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Re: DS - GMATPREP - coordinate geometry -2 [#permalink]

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New post 12 Aug 2008, 11:36
Boss... please type the question..
Sorry --- will do next time

Thanks for the explanation, one question though

intersect x axis when y=0
(x+a)(x+b)=0
--> two points are .. (0,-a) and (0,-b)

Shouldn't the points be (-a,0) and (-b,0)
since x = -a or -b


1) a+b=-1
--> one equation two variable..s

Not suffciient

2) graph intersect y at (0,-6)

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

-6=(0+a) (0+b) --> ab=-6

Not suffcieint

combine both

a+b=-1
ab=-6
a-6/a=-1

a^2-6+a=0
--> a*a+3a-2a-6 =0 --> a(a+3)-2(a+3) ===>
(a+3)(a-2)=0

a=-3 or a=2
substitue values in a+b=-1
a=-3 b=2
a=2 b=-3
(both solutions are same )

suffcieint

C

Kudos [?]: 8 [0], given: 0

SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1795

Kudos [?]: 1032 [0], given: 5

Location: New York
Re: DS - GMATPREP - coordinate geometry -2 [#permalink]

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New post 12 Aug 2008, 11:41
mba9now wrote:
Boss... please type the question..
Sorry --- will do next time

Thanks for the explanation, one question though

intersect x axis when y=0
(x+a)(x+b)=0
--> two points are .. (0,-a) and (0,-b)

Shouldn't the points be (-a,0) and (-b,0)
since x = -a or -b


1) a+b=-1
--> one equation two variable..s

Not suffciient

2) graph intersect y at (0,-6)

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

-6=(0+a) (0+b) --> ab=-6

Not suffcieint

combine both

a+b=-1
ab=-6
a-6/a=-1

a^2-6+a=0
--> a*a+3a-2a-6 =0 --> a(a+3)-2(a+3) ===>
(a+3)(a-2)=0

a=-3 or a=2
substitue values in a+b=-1
a=-3 b=2
a=2 b=-3
(both solutions are same )

suffcieint

C


you are right.. I corrected in the orinal post.
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Re: DS - GMATPREP - coordinate geometry -2   [#permalink] 12 Aug 2008, 11:41
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