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sweetymango
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#1
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How do I solve: 4x^3-3x^4=0.5?
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fishpaste
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Perhaps using an iterative method like Newton Raphson,or you could rearrange it into f(x) = 0 and try and guess factors.
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shift3
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(Original post by fishpaste)
Perhaps using an iterative method like Newton Raphson,or you could rearrange it into f(x) = 0 and try and guess factors.
Hmm, since I still haven't studied the Newton-Raphson method I used another weird (and probably incorrect) method: logarithms.

I ended up with x =~ 1.33, is that right?
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sweetymango
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Well the actual question is from S3:

The Continuous random variable Y has pdf f(y) defined by:
f(y)=12y^2(1-y) for 0<y<1, 0 otherwise
Find the median:
so you want when p<y=0.5
So i did this by integrating 12y^2-12y^3
which is 4y^3-3y^4 = 0.5.
(then someone's going to tell me i've integrated it wrong!)

Due to the nature of question I would have thought there would be an exact answer using a method, other than Newton Raphson, etc etc...
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Squishy
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Yeah, a numerical method would be best (Newton Raphson is fine). Sketch a graph and you'll see there are two solutions.

I get 0.6143 and 1.2475 as my solutions to 4DP.
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john !!
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(Original post by sweetymango)
How do I solve: 4x^3-3x^4=0.5?
Draw the curve 4x^3 - 3x^4 = y and the line y = 0.5, where they cross should be the roots.

There are no nice factors of 8x^3 - 6x^4 - 1 that I can see, so an approximations might be the best you can get from me...
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Trinity Blume
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haha I don't suppose this is the MEI S3 book?
We just started that and it had a few questions like that and we were told not to bother doing them by our teacher because a) they are unsoluable using methods they expect for s3 and b) we'd never get asked such a thing in an exam. In fact our teacher said it probably was an error in printing.
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Squishy
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That is a rather unusual answer for a stats question I must admit. They wouldn't expect you to know Newton-Raphson for an S3 paper.
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sweetymango
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Yeah i understand your methods...i think the book must have done a printing error...or some sort...as it thinks the answer is 2/3
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Jonny W
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(Original post by sweetymango)
Yeah i understand your methods...i think the book must have done a printing error...or some sort...as it thinks the answer is 2/3
2/3 is the mode.
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