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# C1 discriminant gold paper edexcel

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1. https://9d9ee19f35247c2d9536ca2fef66...%20Edexcel.pdf

Help on question 7 please i don't understand
2. (Original post by Big Moisty)
https://9d9ee19f35247c2d9536ca2fef66...%20Edexcel.pdf

Help on question 7 please i don't understand
well what is the discriminant?
3. (Original post by samb1234)
well what is the discriminant?

but i don't know how to do the show bit in the question

4. So

Then complete the square.... for the next bit
5. (Original post by Big Moisty)

but i don't know how to do the show bit in the question
complete the square
6. Just do exactly the same thing as you'd normally do if b and c were regular numbers.
7. (Original post by Math12345)

So

Then complete the square.... for the next bit
ok done it what now???

(Original post by samb1234)
complete the square
OOps sorry all maybe i should've clarified xD

it's part C which i can't do
8. (Original post by Big Moisty)

it's part C which i can't do
If you complete the square you get (squared number) + positive number which is obviously always a positive number, so you've shown that the discriminant is positive for all values of k, what does the discriminant always being >0 mean?
9. (Original post by Zacken)
If you complete the square you get (squared number) + positive number which is obviously always a positive number, so you've shown that the discriminant is positive for all values of k, what does the discriminant always being >0 mean?
It has 2 values? It can't be less than 0?
10. (Original post by Big Moisty)
It has 2 values? It can't be less than 0?
If the discriminant is >0, then there are (two) real roots to the equation f(x) = 0.
11. (Original post by Zacken)
If the discriminant is >0, then there are (two) real roots to the equation f(x) = 0.
Yes but how do i prove that k has values when f(x)=0 ?
12. (Original post by Big Moisty)
Yes but how do i prove that k has values when f(x)=0 ?
What?

Your discriminant is in terms of k.

If you show that your discriminant is always > 0 no matter what k is then you've shown that f(x) = 0 has (two) real roots no matter what.

You show that your discriminant is always > 0 no matter what k is by writing it in the form (squared number) + (positive number) which are both positive terms and hence the discriminant is always positive and > 0.
13. (Original post by Zacken)
What?

Your discriminant is in terms of k.

If you show that your discriminant is always > 0 no matter what k is then you've shown that f(x) = 0 has (two) real roots no matter what.

You show that your discriminant is always > 0 no matter what k is by writing it in the form (squared number) + (positive number) which are both positive terms and hence the discriminant is always positive and > 0.
oops me being derpy.

so do i have to write some words or show some sort of proof using maths?
14. (Original post by Big Moisty)
oops me being derpy.

so do i have to write some words or show some sort of proof using maths?
Just say "since the discriminant is always > 0 then..."
15. (Original post by Zacken)
Just say "since the discriminant is always > 0 then..."
then where the squared bracket is always positive you'll always have a value great than 0?
16. (Original post by Big Moisty)
then where the squared bracket is always positive you'll always have a value great than 0?
Since you have a squared bracket which is always positive (a squared number is always >=0)
+
A positive number which is always positive

Then (squared bracket) + (positive) = positive.
17. (Original post by Zacken)
Since you have a squared bracket which is always positive (a squared number is always >=0)
+
A positive number which is always positive

Then (squared bracket) + (positive) = positive.
ok thanks
18. (Original post by Big Moisty)
then where the squared bracket is always positive you'll always have a value great than 0?
I would just say squared bracket is>= 0 so discriminant is >=positive constant >0 so therefore 2 real roots for all values of k
19. (Original post by samb1234)
I would just say squared bracket is>= 0 so discriminant is >=positive constant >0 so therefore 2 real roots for all values of k
ok thanks
20. (Original post by Big Moisty)
ok thanks
Cool. No problem.

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