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https://917c8db4f385e20466384eff57d816b2bed188db.googledrive.com/host/0B1ZiqBksUHNYMVhvRUNrcWJ6LXM/June%202011%20QP%20-%20C1%20Edexcel.pdf

Question 7c how do you show this?

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Original post by Acrux


What have you tried so far? :h:
Reply 2
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Acrux
OP
7
Original post by SeanFM
What have you tried so far? :h:


solving for k but this gets you no where
Discriminant: (k+3)2−4k=k2+2k+9=(k+1)2+8(k+3)^2-4k = k^2+2k+9 = (k+1)^2+8

Real roots if discriminant ≥0: Clearly (k+1)2+8≥0(k+1)^2+8\geq0 since minimum point is (−1,8)(-1,8) (i.e it lies above the x-axis so is positive)
(edited 7 years ago)
Reply 4
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B_9710
18
f(x) f(x) will have real roots if the discriminant is greater, or equal to 0 (question does not say roots have to be distinct).
Original post by Math12345
Discriminant: (k+3)2−4k=k2+2k+9=(k+1)2+8(k+3)^2-4k = k^2+2k+9 = (k+1)^2+8

Real roots if discriminant ≥0: Clearly (k+1)2+8≥0(k+1)^2+8\geq0 since minimum point is (−1,8)(-1,8)


Well done but would it not be more beneficial to the OP if you questioned them, instead of giving the answer? :smile:
You have to calculate the discriminant.
Original post by Acrux
solving for k but this gets you no where


With questions with multiple parts, most of the time there is a link at some point.

You've found an expression for the discriminant, and you know that when f(x) = 0 and the equation has real roots, the discriminant is...

So the link is..
Original post by The-Spartan
Well done but would it not be more beneficial to the OP if you questioned them, instead of giving the answer? :smile:

Sorry boss
Reply 9
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Acrux
OP
7
Original post by Math12345
Discriminant: (k+3)2−4k=k2+2k+9=(k+1)2+8(k+3)^2-4k = k^2+2k+9 = (k+1)^2+8

Real roots if discriminant ≥0: Clearly (k+1)2+8≥0(k+1)^2+8\geq0 since minimum point is (−1,8)(-1,8) (i.e it lies above the x-axis so is positive)


what does the minimum point imply
so what is positive..?
Original post by Acrux
what does the minimum point imply
so what is positive..?


For f(x)=0 to have real roots, the discriminant must be greater than or equal to 0. Clearly the discriminant (k+1)^2+8 is greater or equal to 0 (sketch it) - I just said the minimum point is (-1,8) to show that the discriminant is ≥0. You could just say (k+1)^2≥0 for the mark (the square of real numbers is positive).
(edited 7 years ago)
Reply 11
Avatar for Acrux
Acrux
OP
7
Original post by Math12345
For f(x)=0 to have real roots, the discriminant must be greater than or equal to 0. Clearly the discriminant (k+1)^2+8 is greater or equal to 0 (sketch it) - I just said the minimum point is (-1,8) to show that the discriminant is ≥0. You could just say (k+1)^2≥0 for the mark (the square of real numbers is positive).


However if the discriminant is greater than zero shouldnt there be roots this graph does not cross the x-axis
Reply 12
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Acrux
OP
7
Original post by SeanFM
With questions with multiple parts, most of the time there is a link at some point.

You've found an expression for the discriminant, and you know that when f(x) = 0 and the equation has real roots, the discriminant is...

So the link is..


How do you workout 9bi?
Original post by Acrux
However if the discriminant is greater than zero shouldnt there be roots this graph does not cross the x-axis


No, you are missing the point. f(x) will cross the x-axis and have roots!, but the discriminant is completely different. We need to show that the discriminant is positive (i.e is always above the x-axis) to show that f(x) has real roots
Original post by Acrux
How do you workout 9bi?


What have you tried so far?
Reply 15
Avatar for Acrux
Acrux
OP
7
Original post by Math12345
No, you are missing the point. f(x) will cross the x-axis and have roots!, but the discriminant is completely different. We need to show that the discriminant is positive (i.e is always above the x-axis) to show that f(x) has real roots


Ok i understand now
Reply 16
Avatar for Acrux
Acrux
OP
7
Original post by SeanFM
What have you tried so far?


Un=k+(100-1)k
=100k
k+2k+3k+....+100=k(1+2+3+...+100k)k+2k+3k+....+100=k(1+2+3+...+ \frac{100}{k})

How many terms?
(edited 7 years ago)
Reply 18
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Acrux
OP
7
Original post by Math12345
k+2k+3k+....+100=k(1+2+...100/k)

How many terms?

100 terms
Original post by Acrux
100 terms


Careful.

Count the bits in the brackets (1,2.....)
(edited 7 years ago)

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