polynomial question (1 Viewer)

[sen00]

Find all real x, such that

x^4 + 2x^3 + 3x^2 + 2x = 10

 

[LoneShadow]

x -> 1/2 (-1 - Sqrt[-3 + 4 Sqrt[11]]);
x -> 1/2 (-1 + Sqrt[-3 + 4 Sqrt[11]]);
x -> 1/2 (-1 - i Sqrt[3 + 4 Sqrt[11]]);
x -> 1/2 (-1 + i Sqrt[3 + 4 Sqrt[11]]).
...from the source of all wisdom: Mathematica




....here's how you do it by hand:

0 = x^4 + 2 x^3 + 3 x^2 + 2 x - 10
= (x^2 + x)^2 + 2 (x^2 + x) - 10
= m^2 + 2m - 10; where m = x^2 + x
so,
m = -1 + sqrt(11) or m = -1 - sqrt(11)

Case m = -1 + sqrt(11):
x^2 + x +1 - sqrt(11) = 0 => x = [-1 + sqrt(-3 + sqrt(11))]/2 or x = [-1 - sqrt(-3 + sqrt(11))]/2

Case m = -1 - sqrt(11):
x^2 + x +1 + sqrt(11) = 0 => x = [-1 + i sqrt(3 + 4 sqrt(11))]/2 or x = [-1 - i sqrt(3 + 4 sqrt(11))]/2


....don't ask useless questions like this no more

 

[sen00]

well it comes down to this:

x^4 + 2x^3 + 3x^2 + 2x + 1 = (x^2 + x + 1)^2

and it pops right out.

pretty neat huh?

i dont think its useless, i think its quite instructive.

i think high school studnets we are too used to squares of linears only and if this was asked, few would be able to solve it in my opinion.

 

[Yip]

Even more neat is the following problem, which uses the above identity

Evaluate the series:
[[1+(1/1^2)+(1/2^2)]^(1/2)]+[[1+(1/2^2)+(1/3^2)]^(1/2)]+...+[[1+(1/2006^2)+(1/2007^2)]^(1/2)]

 

[vulgarfraction]

well it comes down to this:

x^4 + 2x^3 + 3x^2 + 2x + 1 = (x^2 + x + 1)^2

and it pops right out.

pretty neat huh?

i dont think its useless, i think its quite instructive.

i think high school studnets we are too used to squares of linears only and if this was asked, few would be able to solve it in my opinion.

Are you meant to notice that it's symmetric except for the constant, and then bring it onto the other side? Cos then you can put it as a product of symmetric quadratics/linears minus 9, and it works out that you have a square.

Lesson: bash only when you've tried thinking.