username3025666
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Hello,
Could someone help me on how to go about answering the question please.

8) The straight line with equation y = 3x – 7 does not cross or touch the curve with equation y = 2px^2 – 6px + 4p, where p is a constant.

(a) Show that 4p^2 – 20p + 9 < 0

(b) Hence find the set of possible values of p.

Thank you!
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Sir Cumference
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(Original post by Rubydiamond3131)
Hello,
Could someone help me on how to go about answering the question please.

8) The straight line with equation y = 3x – 7 does not cross or touch the curve with equation y = 2px^2 – 6px + 4p, where p is a constant.

(a) Show that 4p^2 – 20p + 9 < 0

(b) Hence find the set of possible values of p.

Thank you!
To find the points of intersection you would solve the equations simultaneously (set the equations equal to each other) to give

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

If you are told that the lines don't intersect then the equation above can't have any solutions. So if the equation has no solutions, what can you say?
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username3025666
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(Original post by notnek)
To find the points of intersection you would solve the equations simultaneously (set the equations equal to each other) to give

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

If you are told that the lines don't intersect then the equation above can't have any solutions. So if the equation has no solutions, what can you say?
Yes, I thought if the equations do not touch/meet, then I would not be able to equate them. However, for part A , the mark scheme has equated the equations?!?

I can link the mark scheme, if that helps...

http://qualifications.pearson.com/co...s_20160817.pdf
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Sir Cumference
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(Original post by Rubydiamond3131)
Yes, I thought if the equations do not touch/meet, then I would not be able to equate them. However, for part A , the mark scheme has equated the equations?!?

I can link the mark scheme, if that helps...

http://qualifications.pearson.com/co...s_20160817.pdf
No I am saying that you should be equating the equations. Then once you've done that you need to show that the equation that you've made has no solutions.

Hint : discriminant.
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username3025666
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(Original post by notnek)
No I am saying that you should be equating the equations. Then once you've done that you need to show that the equation that you've made has no solutions.

Hint : discriminant.
Ah okay, Thank you! I understand now
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