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# Do lines y = ax^2 + b and y = cx^2 + d cross? 1. a = -c 2. b

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Director
Joined: 09 Aug 2006
Posts: 755
Do lines y = ax^2 + b and y = cx^2 + d cross? 1. a = -c 2. b [#permalink]

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04 Jan 2008, 00:27
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Do lines y = ax^2 + b and y = cx^2 + d cross?

1. a = -c
2. b > d

I have no idea about how to approach this question. Can someone please explain? Thanks.

Last edited by GK_Gmat on 08 Jan 2008, 00:24, edited 1 time in total.
CEO
Joined: 29 Mar 2007
Posts: 2560

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04 Jan 2008, 00:52
GK_Gmat wrote:
Do lines y = ax2 + b and y = cx2 + d cross?

1. a = -c
2. b > d

I have no idea about how to approach this question. Can someone please explain? Thanks.

I think its C, but im not 100% sure yet.
Manager
Joined: 01 Jan 2008
Posts: 223
Schools: Booth, Stern, Haas

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04 Jan 2008, 01:04
how can you say that it is C if there is no answers?
Intern
Joined: 27 Oct 2007
Posts: 10

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04 Jan 2008, 04:48
GK_Gmat wrote:
Do lines y = ax2 + b and y = cx2 + d cross?

1. a = -c
2. b > d

I have no idea about how to approach this question. Can someone please explain? Thanks.

Do you mean y=ax^2 + b and y=cx^2 + d?

Ans IMO is E
(BTW these are eqs of parabola)
Director
Joined: 09 Jul 2005
Posts: 591

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04 Jan 2008, 05:58
From S1 we know that a·x^2+b=-a·x^2+d => x= +/- sqrt[(d-b)/(2·a)]
We can not conclude anything because if b=d then the two paraboles cross in just one point so they do not really cross each other.

From S2 we can not conclude anything

From S1 & S2 we know that b and d are different and that x= +/- sqrt[(d-b)/(2·a)] , which means two different points, so they cross.

Director
Joined: 09 Aug 2006
Posts: 755

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08 Jan 2008, 00:25
Anyone who can explain the correct answer??
SVP
Joined: 04 May 2006
Posts: 1894
Schools: CBS, Kellogg

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08 Jan 2008, 01:46
GK_Gmat wrote:
Do lines y = ax^2 + b and y = cx^2 + d cross?

1. a = -c
2. b > d

I have no idea about how to approach this question. Can someone please explain? Thanks.

E, sure

ax^2 +b =cx^2 +d
(a-c)x^2 =d-b
x^2 =(d-b)/(a-c)

the question is the same as whether the final equation has solution?

the equation has solution only when x^2 >=0
1. a=-c, says nothing about (d-b)/(-2c)
2. b>d, the same
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Director
Joined: 09 Jul 2005
Posts: 591

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08 Jan 2008, 06:56
GK_Gmat wrote:
Anyone who can explain the correct answer??

Answer is C. Let me know what is not consistent with my previous explanation.
CEO
Joined: 17 Nov 2007
Posts: 3584
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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08 Jan 2008, 12:11
automan wrote:
From S1 we know that a·x^2+b=-a·x^2+d => x= +/- sqrt[(d-b)/(2·a)]
We can not conclude anything because if b=d then the two paraboles cross in just one point so they do not really cross each other.

From S2 we can not conclude anything

From S1 & S2 we know that b and d are different and that x= +/- sqrt[(d-b)/(2·a)] , which means two different points, so they cross.

(d-b)<0

a>0: (d-b)/(2·a)<0 ==> √((d-b)/(2·a)) is undefined. There are no roots.
a<0: (d-b)/(2·a)>0 ==> √((d-b)/(2·a)) is defined. There are two roots.
Insuff.
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Director
Joined: 09 Jul 2005
Posts: 591

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08 Jan 2008, 15:22
walker wrote:
automan wrote:
From S1 we know that a·x^2+b=-a·x^2+d => x= +/- sqrt[(d-b)/(2·a)]
We can not conclude anything because if b=d then the two paraboles cross in just one point so they do not really cross each other.

From S2 we can not conclude anything

From S1 & S2 we know that b and d are different and that x= +/- sqrt[(d-b)/(2·a)] , which means two different points, so they cross.

(d-b)<0

a>0: (d-b)/(2·a)<0 ==> √((d-b)/(2·a)) is undefined. There are no roots.
a<0: (d-b)/(2·a)>0 ==> √((d-b)/(2·a)) is defined. There are two roots.
Insuff.

Humm...tricky question. I think you are right. I got a bit confused not considering the sign of a. I considered a being either positive or negative. In such a case you can conclude that the paraboles either cross or not cross. You are right, E should be the answer.
Re: DS: Lines   [#permalink] 08 Jan 2008, 15:22
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