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C1 discriminant gold paper edexcel

https://9d9ee19f35247c2d9536ca2fef660ac833b08803.googledrive.com/host/0B1ZiqBksUHNYTzZzRjZEYVQ4TDA/12%20Gold%204%20-%20C1%20Edexcel.pdf

Help on question 7 please i don't understand

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Original post by Big Moisty


well what is the discriminant?
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Original post by samb1234
well what is the discriminant?


b2−4ac b^2 -4ac

but i don't know how to do the show bit in the question
x2+(k+3)x+k[br][br]a=1,b=(k+3)andc=kx^2+(k+3)x+k[br][br]a=1, b=(k+3) and c=k

So b2−4ac=(k+3)2−4(1)(k)=k2+6k+9−4k=k2+2k+9b^2-4ac = (k+3)^2-4(1)(k) = k^2+6k+9-4k = k^2+2k+9

Then complete the square.... for the next bit
Original post by Big Moisty
b2−4ac b^2 -4ac

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


complete the square
Just do exactly the same thing as you'd normally do if b and c were regular numbers.
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Original post by Math12345
x2+(k+3)x+k[br][br]a=1,b=(k+3)andc=kx^2+(k+3)x+k[br][br]a=1, b=(k+3) and c=k

So b2−4ac=(k+3)2−4(1)(k)=k2+6k+9−4k=k2+2k+9b^2-4ac = (k+3)^2-4(1)(k) = k^2+6k+9-4k = k^2+2k+9

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
(edited 7 years ago)
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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?
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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?
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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.
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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 ?
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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.
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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?
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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..."
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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?
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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.
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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
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
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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 :smile:
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Original post by Big Moisty
ok thanks


Cool. No problem.

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