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# Core 1 Graph Question, someone help please? watch

1. so a=1
b = 2
c= 4?
2. (Original post by liverpool2044)
so a=1
b = 2
c= 4?
No. If you plug those into the formula b^2 - 4ac what will you get? Nothing, because there is nothing to solve! You need some k terms, your a is correct however. I'll do "b" for you and you can do c.

b = -(k+2)
3. Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
4. (Original post by liverpool2044)
Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
C is basically any term that isn't multiplied by x. So c = 2k + 1
5. (Original post by liverpool2044)
Ahh i dont know why im so confused by this haha. I suppose theres a 2k left so 2?
a is the sum of the coefficients of x^2
b is the sum of the coefficients of x (x^1)
c is the sum of all the constant terms (you could say coefficients of x^0)
6. Ah okay makes a lot of sense now, thank you.

So its -(k+2)^2-4x1x2k-1

Do i then expand the b^2
7. i got to

k(k-4)+5 so the answers are k = 0 or 4?
8. (Original post by liverpool2044)
Ah okay makes a lot of sense now, thank you.

So its -(k+2)^2-4x1x2k-1

Do i then expand the b^2
Be careful with brackets (also it is 2k+1 not 2k-1)

(-(k+2))^2 - 4*1*(2k+1) = ?

You have to equate this to something. You are looking for a tangent, so how many points of intersection will there be with the curve?
9. (Original post by liverpool2044)
i got to

k(k-4)+5 so the answers are k = 0 or 4?
That's what I got too
10. thanks! great help
11. (Original post by liverpool2044)
thanks! great help
no problem let's see if you understand the b^2 - 4ac now, what's a b and c in each of these:

i) kx^2 - 4x + 3 = 0
ii) 3x - (k+1)x^2 - kx = 0
iii) x^2 - 3(k+1)x + k - 2 = 0

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