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# Does the line represented by y=ax^2+bx+c intersect with

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Does the line represented by y=ax^2+bx+c intersect with [#permalink]  21 Apr 2006, 02:16
Does the line represented by y=ax^2+bx+c intersect with X-axis?
1)a>0
2)c<0
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Re: Co-Ordiante [#permalink]  21 Apr 2006, 10:18
The line would intersect with x asis when y becomes 0, ie, when
ax^2 + bx + c = 0 holds true.

From 1, a > 0.
Thus x^2 is not a zero entity. Thus, the line would have a exponential component in the graph.
Also, x^2 would always be positive, and so ax^2 is a positive entity.

But we don't know the signs of b and c and therefore don't know if this expression turns 0.

From 2, c < 0. Thus, for x = 0, the expression is -ve. If the expression is positive for some values of x, then the expression would also be 0 before turning -ve. but we can't assume that, since we don't know the value of a and b.

Combining, we don't know the value of b. But for big enough values of x, ax^2 would be bigger than bx + c and the expression would be +ve.

Therefore, C.

Can someone solve it in a less cumbersome way?
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Re: Co-Ordiante [#permalink]  21 Apr 2006, 10:29
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!
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Re: Co-Ordiante [#permalink]  21 Apr 2006, 10:40
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.
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Re: Co-Ordiante [#permalink]  21 Apr 2006, 11:38
vivek123 wrote:
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.

Absolutely! I missed that one.
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

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Joined: 14 Dec 2004
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Re: Co-Ordiante [#permalink]  21 Apr 2006, 19:53
giddi77 wrote:
vivek123 wrote:
giddi77 wrote:
kapslock wrote:
Can someone solve it in a less cumbersome way?

Just remember the properties:
If a>0 & c>0 parabola opens upwards and the vertex of the parabola is above x axis and the curve will not intersect the x-axis
If a>0 & c <= 0, the parabola opens upwards and the vertex of parabola is touching x-axis or is below the x-axis. Hence it has one or two points.

The opposite is true for a<0... parabola opens downwards....

You can verify this by putting x=0, If c is -ve, kowning that a>0, the parabola should intersect the x-axis at 2 points!

Girish, one correction.....with a>0 & c>0, the parabola may or may not touch/intersect x-axis

because, possibilities are
b^2 < 4ac,
b^2 = 4ac,
b^2 > 4ac

But with a>0 & c<0, we can definitely say that b^2-4ac > 0, hence intersects x-axis.

Absolutely! I missed that one.

I'm sure that was in sleep mode. Dude, can u lend me your maths skills for a day?
Re: Co-Ordiante   [#permalink] 21 Apr 2006, 19:53
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