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#1 2008-01-30 03:52:26

phamthephong100
Member
Registered: 2007-10-13
Posts: 8

Who can solve this problem???

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#2 2008-01-30 04:33:23

NullRoot
Member
Registered: 2007-11-19
Posts: 162

Re: Who can solve this problem???

It factors out to:
(2x - 1)(x²  - 2x + 3)

X can be 0.5 so that (2x-1) = 0

For (x²  - 2x + 3) = 0, you need to go into complex numbers. I think the answer's something like 1 ± i√2


Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

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#3 2008-01-30 04:46:50

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Who can solve this problem???

x²  - 2x + 3 = (x-1)²+2
x-1 = ±i√2
x = 1±i√2


The Beginning Of All Things To End.
The End Of All Things To Come.

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#4 2008-01-30 05:08:54

NullRoot
Member
Registered: 2007-11-19
Posts: 162

Re: Who can solve this problem???

Thanks, Luca. I thought that was right, but wasn't sure. I knew there weren't any real answers, though wink


Trillian: Five to one against and falling. Four to one against and falling… Three to one, two, one. Probability factor of one to one. We have normality. I repeat, we have normality. Anything you still can’t cope with is therefore your own problem.

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#5 2008-01-30 11:16:21

LuisRodg
Real Member
Registered: 2007-10-23
Posts: 322

Re: Who can solve this problem???

After looking at the equation:

2x^3-5x^2+8x-3=0

How would one know that it factors out to:

2x - 1)(x²  - 2x + 3)

I mean, after having the factors, it is obvious but starting with the equation what are the steps to figure out the factors?

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#6 2008-01-30 15:28:03

Identity
Member
Registered: 2007-04-18
Posts: 934

Re: Who can solve this problem???

phamthephong100 wrote:

Here's one way...

There can be one linear and one quadratic factor, so

So we are left with

Solving these simultaneously gives

,
,
.

So

In fact, that's the only way I can think of, unless you are a genius with algebraic manipulation or you get lucky with the factor theorem.

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