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

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

Question 7c how do you show this?
What have you tried so far?
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Acrux
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(Original post by SeanFM)
What have you tried so far?
solving for k but this gets you no where
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Math12345
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Discriminant: (k+3)^2-4k = k^2+2k+9 = (k+1)^2+8

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

Real roots if discriminant ≥0: Clearly (k+1)^2+8\geq0 since minimum point is (-1,8)
Well done but would it not be more beneficial to the OP if you questioned them, instead of giving the answer?
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drinktheoceans
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You have to calculate the discriminant.
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Kevin De Bruyne
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(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..
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Math12345
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(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
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Acrux
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(Original post by Math12345)
Discriminant: (k+3)^2-4k = k^2+2k+9 = (k+1)^2+8

Real roots if discriminant ≥0: Clearly (k+1)^2+8\geq0 since minimum point is (-1,8) (i.e it lies above the x-axis so is positive)
what does the minimum point imply
so what is positive..?
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Math12345
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(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).
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Acrux
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(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
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Acrux
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(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?
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Math12345
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(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
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Kevin De Bruyne
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(Original post by Acrux)
How do you workout 9bi?
What have you tried so far?
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Acrux
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(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
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Acrux
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(Original post by SeanFM)
What have you tried so far?
Un=k+(100-1)k
=100k
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Math12345
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k+2k+3k+....+100=k(1+2+3+...+ \frac{100}{k})

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

How many terms?
100 terms
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Math12345
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(Original post by Acrux)
100 terms
Careful.

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