Maths Difficulty.....Please Help Somebody :)
Maths and statistics discussion, revision, exam and homework help.
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Theone could you just confirm that:(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.
(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.(Original post by Bhaal85)
Theone could you just confirm that:
(x) = 3[(x-1r1r3)^2+10r2r3] in completed square form
is the right answer.
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
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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*
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dy/dx = 6x+8 right.(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
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)
