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Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
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

is the right answer.
2. just draw it and look
3. (Original post by clever_lad)
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

is the right answer.
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...
9. Sorry i've made and error it should read

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.
11. (Original post by clever_lad)
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)
Sorry i've made and error it should read

3((x+4/3)^2-32/9) i think.
10r2r3 not 32/9
15. (Original post by cimmie)
i was right!!
hmmpf whats about funny pictures??
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
i had it first...
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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