What are the real roots of: x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0?

Question

Bonus if you can find them all.
Answers must be exact.
Approximations will not be considered for best answer.

To reiterate: no approximations.

Anonymous, you may have no clue, but if you had Mathematica you could do just as well as the others so far...In fact, I'd be willing to bet that some first graders could...

can_t_get_enough : fix your link
math_kp : fix your arithmetic

Note: the first two answerers did give correct approximations. Use those to verify exact ones.

Answer 1

x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0
x^6 - 6x^5 + 15x^4 - 20x^3 + 15x^2 - 6x + 1 - 10x^3 = 0
So (x - 1)^6 - 10x^3 = 0
So ((x - 1)²)³ - (³√10x)³ = 0

So ((x - 1)² - ³√10x)((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0

So ((x - 1)^4 + ³√10x((x - 1)² + ³√100x²) = 0 N0 real roots (Min value near x = 0.2 just more than 0.87)

OR (x - 1)² - ³√10x = 0
ie x² - (2 + ³√10)x + 1 = 0
ie x = ½[(2 + ³√10) ± √(4 + 4³√10 + ³√100 - 4)
= ½[(2 + ³√10) ± √(4³√10 + ³√100)]

SOLUTION
x = ½[(2 + ³√10) ± √(4³√10 + ³√100)]

Answer 2

The eqution is very close to (x-1)^6 except coefficient of x^3 which is -30 x^3 instead of -20 x^3.

so equation is

(x-1)^6 - 10x^3 = 0

let 10^(1/3) -= t

so we get (x-1)^6 - (tx)^3 = 0

x is not zero

so devide by x^3
(x-1/x)^3 = t

(x-1/x) = t or tw or tw^2 where w = cube root of 1

x - 1/x = t

means x^2-tx + 1 = 0

add and subrtact t^2/4

we get (x-t/2)^2 = t^2/4-1

x = t/2 +/-(t^2/4-1)^(/1/2)

putting t = 10^(1/3) we get 2 solutions

multiplying above 2 by w and w^2 we get 2 + 2 more solutions this is becuase sqrt(w) = w^2 and sqrt(w)^2 = w

x1 = (10^(1/3)) + sqrt(100^(1/3)/4-1)
x2 = (10^(1/3)) - sqrt(100^(1/3)/4 -1)
X3 = x1 cis pi/3
x4 = x2 cis pi/3
x5 = x1 cis 2pi/3
x6= x2 cis 2pi/3

As obvious x1 and x2 are real roots and x3 to x6 are complex roots

Answer 3

x^6 - 6x^5 + 15x^4 - 30x^3 + 15x^2 - 6x + 1 = 0
<-> (x - 1)^6 - 10x^3 = 0
<-> (x - 1)^6 = 10x^3
<-> (x-1)^2 = x.10^(1/3)
<-> x^2 - [2 + 10^(1/3)]x + 1 =0
This equality has 2 real root.
Click link below to see my pic of solution:
msquare.kaist.ac.kr/cgi-bin/mimetex.cgi?(x-1)^6=10x^3%20%20%3C-%3E%20%20(x-1)^2=x.\sqrt[3]{10}%20%20%3C-%3E%20%20x^2-(2+\sqrt[3]{10})x+1=0

Answer 4

there is not any confusion. it is ordinary. -9x^2-6x-a million=0 multiply by -a million the completed equation 9x^2+6x+a million=0 9x^2+3x+3x+a million=0 3x[3x+a million]+a million[3x+a million]=0 [3x+a million][3x+a million]=0 [3x+a million]^2=0 x=-a million/3,-a million/3 the roots are actual,rational and equivalent because of the fact the expression is a desirable sq.

Answer 5

Probalby not the answer you wanted, try sticking it into a graphing calc.
I used mathematica..took about 3 seconds
0.256549
3.89789
The rest are not real

Answer 6

Answers with matlab

x= 0.25655
or
x= 3.89789
or
x= 0.137175(+ or -)0.394706*i
or
x= 0.785608(+ or -)2.260502*i

Elapsed calculation time is 4.786366 seconds.

Sorry, but cant workout steps for you.

All the best

Answer 7

I have no clue....I'm only 12.