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Naz1
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#1
Report Thread starter 16 years ago
#1
how do i make x the subject in this formula:

V = 4XQUBED-(2L-2W)X SQUARD +XLW/
V EQUALS FOUR EX DOUBLED MINUS BRAKET TWO EL MINUS TWO DOUBLEU BRAKET CLOSED EX SQUARD PLUS EX EL W

PLEASE HELP
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Baron
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#2
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#2
wtf is dubed
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elpaw
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#3
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(Original post by Naz1)
how do i make x the subject in this formula:

V = 4XQUBED-(2L-2W)X SQUARD +XLW/
V EQUALS FOUR EX DUBED MINUS BRAKET TWO EL MINUS TWO DOUBLEU BRAKET CLOSED EX SQUARD PLUS EX EL W

PLEASE HELP
V = 4 x³ - (2l-2w) x² + xlw

It is quite difficult to make x the subject, you need to use the 3rd orde equivalent of the "quadratic" formula, which has one real answer (which is probably the one you want, seeing as the equation looks physical)
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john !!
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wrong. can't be bothered to try again.
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john !!
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#5
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(Original post by elpaw)
V = 4 x³ - (2l-2w) x² + xlw

It is quite difficult to make x the subject, you need to use the 3rd orde equivalent of the "quadratic" formula, which has one real answer (which is probably the one you want, seeing as the equation looks physical)
What is that formula? Do you know that one exists?
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elpaw
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(Original post by mik1a)
What is that formula? Do you know that one exists?
yes i do know it exists. i dont know the formula off the top of my head
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chrisbphd
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It is, errm, rather difficult to do that any other way apart from using the formula. I suspect that using the cubic formula will throw out some horrible answer though.
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elpaw
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#8
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(Original post by mik1a)
V/x = (2x - L)(2x + w)
= 4x² + (2w + 2L)x + Lw

use the quadratic formula:

a = 4
b = (2w - 2L)
c = Lw

x = ( -b +/- sqroot (b² - 4ac) ) / 2a
= 2L - 2w +/- sqroot ( (2w-2L)² - 4*4*Lw) / 2*4
= 2L - 2w +/- sqroot ( 4w² + 4L² - 24Lw ) / 8

more or less
you cant use the quad formula, because the RHS has to be 0, and yours is V/x
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chrisbphd
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And when you multiply out the brackets the constant term has to be +LW not -LW (as your factorisation suggests). All in all, that's a bad piece of maths.
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JamesF
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#10
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#10
(Original post by mik1a)
What is that formula? Do you know that one exists?
There are general formulas for cubics and quartics, but not for any higher order equations.
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chrisbphd
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#11
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Here's the formula - it's not nice.
Attached files
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elpaw
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#12
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(Original post by chrisbphd)
It is, errm, rather difficult to do that any other way apart from using the formula. I suspect that using the cubic formula will throw out some horrible answer though.
ive used mathematica and it's a horrible answer. im trying to format it and post it. (and thats only the real answer! the imaginaries are worse!)
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chrisbphd
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#13
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(Original post by chrisbphd)
Here's the formula - it's not nice.
Sorry, I should add that a is the coefficient of x cubed, b is the coefficient of x squared, c is the coefficient of x and d is the constant term. It is only valid for a(x^3)+b(x^2)+cx+d=0
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Sahir
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#14
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(Original post by chrisbphd)
Here's the formula - it's not nice.
uh-oh spaghetti o's
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lgs98jonee
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#15
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#15
(Original post by Naz1)
how do i make x the subject in this formula:

V = 4XQUBED-(2L-2W)X SQUARD +XLW/
V EQUALS FOUR EX DUBED MINUS BRAKET TWO EL MINUS TWO DOUBLEU BRAKET CLOSED EX SQUARD PLUS EX EL W

PLEASE HELP
i assume this is the coursework for finding max vol of box so u want to first differentiate and the max for cetain values of l and w will be when it equals 0
differentiate:
v/x=12x^2-(2l-2w)+LW
when this equals 0,
however, i think u may have gone wrong sumwhere

but u can use quad formula to find values of x for certain ls and ws
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Naz1
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#16
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(Original post by elpaw)
you cant use the quad formula, because the RHS has to be 0, and yours is V/x
Thank you
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username9816
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#17
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#17
(Original post by chrisbphd)
Here's the formula - it's not nice.
*shudders*

Who on earth came up with that!
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elpaw
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#18
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mathematica gives the real answer as:

x = 2(l-w)/3 + (2^(1/3))*(-4(l -w)²+3lw)/(3(-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³ + √(4(-4(l-w)²+ 3 lw)³ + (-16 l³ - 27V + 66 l²w - 66 lw² +16 w³)²))^(1/3)) - (1/(3*2^(1/3))) ((-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³ + √(4(-4(l - w)² + 3 lw)³ + (-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³)²))^(1/3))))
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john !!
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#19
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#19
(Original post by chrisbphd)
Here's the formula - it's not nice.
I wonder how you'd prove that. Completing the cube?
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Naz1
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#20
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#20
(Original post by elpaw)
mathematica gives the real answer as:

x = 2(l-w)/3 + (2^(1/3))*(-4(l -w)²+3lw)/(3(-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³ + √(4(-4(l-w)²+ 3 lw)³ + (-16 l³ - 27V + 66 l²w - 66 lw² +16 w³)²))^(1/3)) - (1/(3*2^(1/3))) ((-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³ + √(4(-4(l - w)² + 3 lw)³ + (-16 l³ - 27 V + 66 l²w - 66 lw² + 16 w³)²))^(1/3))))

Are you sure thats the only way i can make X the subject & how about this one:
SIMPLIFY IT, THEN MAKE X THE SUBJECT
V=(L-2X)*(W-2X)*X
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