Solve for x!

[SilverSprite]

Solve for x

1.(x+1)^5= x^5+1
2.(x+1/x)^2-5(x+1/x)=6

[Kalkewl8ter]

(x+1)^5=x^5+1
x^5+5x^4+10x^3+10x^2+5x+1=x^5+1
5x(x^3+2x^2+2x+1)=0
x(x+1)(x^2+x+1)=0
x=0, or x=-1, or x=-1/2+/- 3i

[sql_lall]

I'm not exactly sure, but i think that (1) and (2) are different Questions.

For 2):
**(x <> 0, as there is the term 1/x)

let y = (x+1/x) = (x² + 1)/x
_________(from this x²-y(x) + 1 =0)
=> y² - 5y - 6 = 0
=> (y-6)(y+1) = 0
=> y = 6 or -1

=> either
x²-6x + 1 =0 -(2 answers)
or
x²+x + 1 =0 -(2 answers)

(Both are simple application of quadratic forumla)
4 answers -none are 0

[Guv]

Assuming that there are two different problems, the solution to the first problem by Kalkewl8ter looks good to me.

There seems to be some confusion in the Sql_Lall post about the second problem. As I understand his analysis, he correctly solves for a variable Y, which can be 6 or minus one.

Y = (x+1)/x (defined as such early in the post).

Y = 6 implies x = 1/5
Y = -1 implies x = -1/2

The above values plugged into the second equation seem to work okay.

[SilverSprite]

The first questions asked for solutions in the complex domain! I'm sorry if i missed that part!

[DiGiTaIErRoR]

Oh, you're asking seperate? I thought you meant solve for x applicable to both equations....

otherwise

1) x= -1/2 (+/-) sqrt(3)i/2 or x = 0 or x =-1

2)x= -1/2 (+/-) sqrt(3)i/2 or

x = (+/-) (2*sqrt(2)(+/-)3) where (+/-) both are + or both are -

unless by -5 you mean multiply and not subtract?

if you meant multiplied by -5 then:

2) x=.379455070524 (+/-) 1.53443682794i or
x= .151874214067 (+/-) .614147511474i or
x= (+/-)(sqrt(6^(2/3)*5^(4/3)-100) (+/-) 6^(1/3)*5^(2/3)) / 10 where (+/-) are inverse

And I wasn't thinking imaginary... so even if you were asking what I originally thought....

x=-1/2 (+/-) sqrt(3)i/2

[sql_lall]

There seems to be some confusion in the Sql_lall post about the second problem. As I understand his analysis, he correctly solves for a variable Y, which can be 6 or minus one.

Y = (x+1)/x (defined as such early in the post).

Y = 6 implies x = 1/5
Y = -1 implies x = -1/2

The above values plugged into the second equation seem to work okay.

Actually, y = x +(1/x), or as i put it: y = (x²+1)/x
i think there is some confusion over whether
x+1/x = (x+1)/x
OR = x+(1/x)
I took it as the latter.

[Guv]

Sql_Lall: After rereading the original post by SilverSprite, I think your interpretation is correct: x+1/x, not (x+1)/x. I did not read it properly.

Hence your post is correct, and solving the quadratics you developed would give correct solutions.