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1. (Original post by theone)
Sure, all quadratics can be written in the form x^2 +bx + c.

Now we can note by multiplying out that this is in fact equal to (x+(b/2))^2 + (c - b^2/4).

This procedure of getting the second equation from the first is called completing the square.

By extending this you can work out the more general formula for the solution of a quadratic.
Theone could you just confirm that:

(x) = 3[(x-1r1r3)^2+10r2r3] in completed square form

2. just draw it and look
i am correct, you dont ahve a clue what you are talking about
No, you've forgotten to divide the 8x (the 4) by 3 when you took it out... it should be 4/3.
4. (Original post by Bhaal85)
Theone could you just confirm that:

(x) = 3[(x-1r1r3)^2+10r2r3] in completed square form

Well, it ain't going to be since you have - 4/3 and the original equation has a positive term in x.

I get f(x) = 3((x+4/3)^2 - 16/9) which gives x = -4/3 as the max.

You get the same by differentiation
5. (Original post by ++Hex++)
No, you've forgotten to divide the 8x (the 4) by 3 when you took it out... it should be 4/3.
So I am right. And that arse was just being rude and arrogant?
6. (Original post by theone)
Well, it ain't going to be since you have - 4/3 and the original equation has a positive term in x.

I get f(x) = 3((x+4/3)^2 - 16/9) which gives x = -4/3 as the max.

You get the same by differentiation
*goes and gets pen and paper*
7. (Original post by Bhaal85)
So I am right. And that arse was just being rude and arrogant?
Who's the arse?
8. (Original post by Bhaal85)
So I am right. And that arse was just being rude and arrogant?
No need for that. Everyone makes mistakes. Especially where maths is involved...

3((x+4/3)^2-32/9) i think.
10. (Original post by theone)
Who's the arse?

Yeah your right dy/dx=0 at x=-8/6. ben revising for Discrete for 4 weeks. lol. Not arsed, but that arse was wrong.
i am correct, you dont ahve a clue what you are talking about
Thats what the arse wrote, to m that is being rude and arrogant, considering I was helping him.
12. i was right!!
13. (Original post by Bhaal85)
Thats what the arse wrote, to m that is being rude and arrogant, considering I was helping him.
Oh, ok, thought that you were having a go at me, no offence
14. (Original post by theone)

3((x+4/3)^2-32/9) i think.
10r2r3 not 32/9
15. (Original post by cimmie)
i was right!!
16. (Original post by cimmie)
i was right!!
how are you right??? Hmm...you must be notsoclever_lad then???
17. (Original post by Bhaal85)
10r2r3 not 32/9
No i don't think.... since we want to complete the square of x^2 +8x/3 +16/3 = (x+4/3)^2 + 16/3 - 16/9.

And 16/3 - 16/9 = 32/9
18. oops, sorry Bhall, i didnt complete the square correctly. i am really sorry mate, i take my comment back *shrieks with embarrasment*........if you dont mind mate, please could you go through the process of completing the square for this equation, then showing me what x value will give the minimum value for the parabola.....please mate...once again, sorry. *shudders*
19. (Original post by cimmie)
x=-4/3

f'(x):6x+8
=>6x+8=0
<=>x=-4/3

sorry its just guessed
but forget this stupid thing! Its just confusing me!
20. (Original post by theone)
No i don't think.... since we want to complete the square of x^2 +8x/3 +16/3 = (x+4/3)^2 + 16/3 - 16/9.

And 16/3 - 16/9 = 32/9
dy/dx = 6x+8 right.

set to equal 0 = 6x+8=0
rearrange to make x subject you get: x=-8/6

sub that into orignal equation 3x^2+8x+16 = 10r2r3. (ten and two thirds)

Updated: October 28, 2003
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