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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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Do lines y = ax^2 + b and y = cx^2 + d cross? 1. a = -c 2. b [#permalink]  04 Sep 2008, 04:47
Do lines $$y = ax^2 + b$$ and $$y = cx^2 + d$$ cross?

1. $$a = -c$$
2. $$b \gt d$$

*
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 04:59
I vote for E

A. a=-c

y=-cx^2+b
y=cx^2+d

Solving for y gives, y=(b+d)/2

Solving for x gives, x^2=(b-d)/2c.... Insuff

B. b>d, (b-d)>0.... Insuff
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 05:12
IMO E.

But those equations are not lines.
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 06:11
arjtryarjtry wrote:
Do lines $$y = ax^2 + b$$ and $$y = cx^2 + d$$ cross?

1. $$a = -c$$
2. $$b \gt d$$

*

These two equations represent parabolas.

1) y = ax^2 + b
y = -ax^2 + d
y=(b+d)/2 ax^2= ((b+d)/2 -b)) = d-b/2
x^2=(d-b)/2a
we don't about a, b, d whether they are positive , negative..
X^2 can have multiple solutions or no real solution (when (d-b)/2a is negative)

When x^2 has real solution or solutions. --> two parabolas cross each other.
When x^2 has no real solution or solutions. --> two parabolas do not cross each other.
2) b>d
insuffcient. multiple solutions possible or no solution possible.

combined.
we know that (d-b) is always negative. but "a" can be -ve or +ve
x^2=(d-b)/2a
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 06:21
I answered by visualising the two parabolas:

We have two parabolas. From Statement 1, we know that one of them is facing downward, the other - upward. Not sufficitent. From Statement 2, we know that the apex (apex, origin, starting point, how is this called??) of the first parabola is placed higher than that of the second. Again not sufficient.

Combine the two statements: we know that the parabolas are either facing each other, or that they are facing in opposite directions, while one of them is placed lower on the y axis than the other. Not sufficient.
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 06:28
Nerdboy wrote:
I answered by visualising the two parabolas:

We have two parabolas. From Statement 1, we know that one of them is facing downward, the other - upward. Not sufficitent. From Statement 2, we know that the apex (apex, origin, starting point, how is this called??) of the first parabola is placed higher than that of the second. Again not sufficient.

Combine the two statements: we know that the parabolas are either facing each other, or that they are facing in opposite directions, while one of them is placed lower on the y axis than the other. Not sufficient.

Good..
I went through both approaches.. (Visualisation is best way.... but difficult to post here.. )
got E both ways.
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 06:41
Guys,

I freaked out when I read the word lines for y=ax ^2 +b.

It came down to C & E (POE) and I picked E.

Should we be prepared to an extent where we get deep into parabola's and beyond?
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Re: LINES CROSS? [#permalink]  04 Sep 2008, 08:01
I think A,
lines y = ax^2 + b and y = cx^2 + d cross when the system of equations y = ax^2 + b and y = cx^2 + d has any solution.
So with every x >>> y = (b+d)/2 with every b and d.
Since the the system of equations y = ax^2 + b and y = cx^2 + d has always at least a solution, lines y = ax^2 + b and y = cx^2 + d cross
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Re: LINES CROSS? [#permalink]  17 Sep 2011, 01:44
Here are the two cases.
For set of equations y=-x^2+1 and y=x^2, parabolas will intersect.
For set of equations y=x^2+1 and y=-x^2, parabolas will not intersect
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Re: LINES CROSS? [#permalink]  17 Sep 2011, 17:57
fiesta wrote:
I think A,
lines y = ax^2 + b and y = cx^2 + d cross when the system of equations y = ax^2 + b and y = cx^2 + d has any solution.
So with every x >>> y = (b+d)/2 with every b and d.
Since the the system of equations y = ax^2 + b and y = cx^2 + d has always at least a solution, lines y = ax^2 + b and y = cx^2 + d cross

You are right about value of y = (b+d)/2.

However, solution of the system of equations y = ax^2 + b and y = cx^2 + d is not only y, but x.

With y= (b+d)/2, (b+d)/2= ax^2+b
==> ax^2= (d-b)/2
If a and (d-b) have different signs, no solution can be found.

Hence, I think E
Re: LINES CROSS?   [#permalink] 17 Sep 2011, 17:57
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