Solve for x
1.(x+1)^5= x^5+1
2.(x+1/x)^2-5(x+1/x)=6
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Solve for x
1.(x+1)^5= x^5+1
2.(x+1/x)^2-5(x+1/x)=6
(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
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
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.
The first questions asked for solutions in the complex domain! I'm sorry if i missed that part!
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
Actually, y = x +(1/x), or as i put it: y = (x2+1)/xQuote:
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.
i think there is some confusion over whether
x+1/x = (x+1)/x
OR = x+(1/x)
I took it as the latter.
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.