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Do lines \(y = ax^2 + b\) and \(y = cx^2 + d\) cross?

1. \(a = -c\) 2. \(b \gt d\)

*

These two equations represent parabolas.

IMO E is the answer.

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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Your attitude determines your altitude Smiling wins more friends than frowning

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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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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Your attitude determines your altitude Smiling wins more friends than frowning

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

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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My dad once said to me: Son, nothing succeeds like success.

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.

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