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c1 question help

Hi can someone help me, I would of had an attempt but I have no idea where to begin. Is this discriminant question? Help would be appreciated
Original post by junayd1998
Hi can someone help me, I would of had an attempt but I have no idea where to begin. Is this discriminant question? Help would be appreciated


Start by equating the two equations, they don't intersect, so the quadratic would have no real solutions.
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junayd1998
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Original post by NotNotBatman
Start by equating the two equations, they don't intersect, so the quadratic would have no real solutions.


huh equating the equations? Sorry I'm confused.
Original post by junayd1998
huh equating the equations? Sorry I'm confused.


y = 3x-7 and y= 2px^2 -6px + 4p.

so 3x-7 = 2px^2-6px +4p

You know have a quadratic equation, the next steps are standard.
Original post by junayd1998
huh equating the equations? Sorry I'm confused.


lets say if the equations had solutions (if they intersected) then you would rearrange for something like y in both equations and equate them. When you equate them you will get a quadratic. Now if a quadratic has real solutions b^2-4ac>0 and if it has no real solutions it will be less then 0. So when you basically equate them you need to show it has no real solutions
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Original post by Asad_2015
lets say if the equations had solutions (if they intersected) then you would rearrange for something like y in both equations and equate them. When you equate them you will get a quadratic. Now if a quadratic has real solutions b^2-4ac>0 and if it has no real solutions it will be less then 0. So when you basically equate them you need to show it has no real solutions


so do we rearrange 3x-7=2px^2-6px+4p ?

into a quadratic ?
Original post by junayd1998
so do we rearrange 3x-7=2px^2-6px+4p ?

into a quadratic ?


When you have a quadratic equation you make it equal to 0.
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Original post by NotNotBatman
When you have a quadratic equation you make it equal to 0.


yeah so, 2px^2-6px+4p+7-3x = 0 ?
Original post by junayd1998
yeah so, 2px^2-6px+4p+7-3x = 0 ?


Try grouping the x terms.
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Original post by NotNotBatman
Try grouping the x terms.


There are no like terms I'm confused.
Original post by junayd1998
There are no like terms I'm confused.


It would be easier to have all the x terms together, so we can say 2px^2 + (-6p-3)x + 4p+7 = 0
Writing it like this makes it easier to see what the a,b and c values are.
Now for no points of intersection, there are no real solutions so b^2-4ac < 0
Original post by NotNotBatman
It would be easier to have all the x terms together, so we can say 2px^2 + (-6p-3)x + 4p+7 = 0
Writing it like this makes it easier to see what the a,b and c values are.
Now for no points of intersection, there are no real solutions so b^2-4ac < 0


ohhh okay yeah i get it thanks :smile:

and then for part b we just solve the quadratic equation given in part a ?
Original post by junayd1998
ohhh okay yeah i get it thanks :smile:

and then for part b we just solve the quadratic equation given in part a ?


The quadratic inequality, yes.
Original post by NotNotBatman
The quadratic inequality, yes.


So part B would be 1/2<x<18/4?
How did you access this years maths paper? (Just out of curiousity)
Original post by junayd1998
So part B would be 1/2<x<18/4?


Yes, 18/4 can be simplified as 9/2.

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