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#1 2006-01-14 23:56:45

krisper
Member
Registered: 2005-12-20
Posts: 19

Looks simple but actually quite hard

Hi guys,
I have a problem with an equation. Here it is: (x^2 + 4x + 5)/(x+1) = 3√(x+2). I have tried everything but still not getting the roots?! Please help. Thanks.

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#2 2006-01-15 02:39:01

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Looks simple but actually quite hard

Square both sides, mutiply by the denominator, then subtract.  You end up with:

x^4 - x^3 - 10x^2 - 5x + 7 = 0

Which factors into:

(x^2 + x - 1)(x^2 - 2x - 7) = 0

And you use the quadratic from there.  It comes out to (-√5 + 1)/2, (√5- 1)/2, -2√2 + 1, and 2√2 - 1.

Last edited by Ricky (2006-01-15 02:39:32)

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-01-15 02:43:57

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Looks simple but actually quite hard

How on earth did you manage to factor that?
Was it trial and error, or was there some obscure method that I haven't heard of?

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It wanted to be normal.

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#4 2006-01-15 04:15:04

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Looks simple but actually quite hard

I'm just that good.

Nah, I used a ti-89.  It uses some fairly complex algorithms to do it, most of which would be unusable by a human.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-01-15 05:25:35

krisper
Member
Registered: 2005-12-20
Posts: 19

Re: Looks simple but actually quite hard

Ricky,

Can you show us the algorithm that is being used or write here how  x^4 - x^3 - 10x^2 - 5x + 7 = 0 is being factored as it is very important for me to understand it. Thanks.

Humankind's inherent sense of right and wrong cannot be biologically explained.

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#6 2006-01-15 05:50:36

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Looks simple but actually quite hard

Boy, if only I could.  But Texas Instruments protects stuff like that.

But it isn't krisper.  Factoring a quartic equation is only an exercise if you are bored or insane.  No math teacher woudl ever have you do this, nor would you learn anything from doing it.

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#7 2006-01-15 06:36:30

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: Looks simple but actually quite hard

We must solve the system
cd=7
a+b=-1
ab+d+c=-10

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#8 2006-01-15 06:40:18

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

Re: Looks simple but actually quite hard

And one more condition: