Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

|
Options

krassi_holmz
2006-01-16 05:40:18

And one more condition:

krassi_holmz
2006-01-16 05:36:30

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

Ricky
2006-01-16 04:50:36

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.

krisper
2006-01-16 04:25:35

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.

Ricky
2006-01-16 03:15:04

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.

mathsyperson
2006-01-16 01:43:57

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?

Ricky
2006-01-16 01:39:01

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.

krisper
2006-01-15 22:56:45

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.