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1. https://917c8db4f385e20466384eff57d8...%20Edexcel.pdf

Question 7c how do you show this?
2. (Original post by Acrux)
https://917c8db4f385e20466384eff57d8...%20Edexcel.pdf

Question 7c how do you show this?
What have you tried so far?
3. (Original post by SeanFM)
What have you tried so far?
solving for k but this gets you no where
4. Discriminant:

Real roots if discriminant ≥0: Clearly since minimum point is (i.e it lies above the x-axis so is positive)
5. will have real roots if the discriminant is greater, or equal to 0 (question does not say roots have to be distinct).
6. (Original post by Math12345)
Discriminant:

Real roots if discriminant ≥0: Clearly since minimum point is
Well done but would it not be more beneficial to the OP if you questioned them, instead of giving the answer?
7. You have to calculate the discriminant.
8. (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...

9. (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?
Sorry boss
10. (Original post by Math12345)
Discriminant:

Real roots if discriminant ≥0: Clearly since minimum point is (i.e it lies above the x-axis so is positive)
what does the minimum point imply
so what is positive..?
11. (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).
12. (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
13. (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...

How do you workout 9bi?
14. (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
15. (Original post by Acrux)
How do you workout 9bi?
What have you tried so far?
16. (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
17. (Original post by SeanFM)
What have you tried so far?
Un=k+(100-1)k
=100k

18. How many terms?
19. (Original post by Math12345)
k+2k+3k+....+100=k(1+2+...100/k)

How many terms?
100 terms
20. (Original post by Acrux)
100 terms
Careful.

Count the bits in the brackets (1,2.....)

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