Question

roots of a polynomial

Answer 1

let the roots of the quadratic equation \[x^2 + x -1 = 0\] be \[\alpha ~~ and ~~ \beta\]

find the value of \[\alpha^6 + \beta^6\]

Answer 2

:| can we simply find roots

Answer 3

its mathematics... you can do what you like as long as it makes sense

Answer 4

It's one negative sign away from the golden ratio's formula. :,(

Answer 5

haha

Answer 6

i found these at a 2nd hand book sale... lots of fabulous questions.

Answer 7

http://www.wolframalpha.com/input/?i=%281%2F2%28sqrt5-1%29+%29%5E6%2B+%281%2F2%28-sqrt5-1%29%29%5E6

Answer 8

hmm

Answer 9

Vieta's formula and Newton sums are our best friends here

Answer 10

aww... I'm disappointed... its easier than you think

Answer 11

this is a high school book.... so think high school maths

Answer 12

(alpha)^2 + (beta)^2 =3 mmm

Answer 13

start with \[a^2 + b^2 = (a+b)^2 - 2ab\]

Answer 14

\[\Large \alpha^2+\alpha = 1 \\ \Large \beta^2 + \beta = 1 \\ \Large \alpha^2 +\alpha = \beta^2 + \beta \\ \Large \alpha^2 - \beta^2 = \beta - \alpha \\ \Large (\alpha + \beta) (\alpha - \beta) = - (\alpha-\beta) \\ \Large \alpha + \beta = -1\]

Answer 15

hehe

Answer 16

to check your answer
\[
\alpha ^6+\beta ^6=18
\]

Answer 17

\[a^6 + b^6 = (a^2)^3+(b^2)^3 = (a^2+b^2)(a^4 -a^2b^2 +b^4)\]

Answer 18

rest should be easy but i think newton sums is more neat

Answer 19

post more @campbell_st please :)

Answer 20

as long as its only algebraic its hmm

Answer 21

thats correct @ganeshie8 and well done @eliassaab 18 is the answer... and use the information from the quadatic\[\alpha + \beta = -1~~~~\alpha \times \beta = -1\]

Answer 22

Interesting, I am not sure I know why those are true, but you can say that if this was the characteristic equation of a 2x2 matrix, then the trace would be -1 and determinant would be -1.

Answer 23

What if we approximate it with Newton Raphson Method. \[x _{i+1} = x _{i} - \frac{ f(x _{i)} }{ f'(x _{i)} }\]

Answer 24

Then eventually it will converge to the root.

Answer 25

@ kainui for a quadratic \[ax^2 + bx + c = 0\] the roots \[\alpha + \beta = \frac{-b}{a}~~~\alpha \times \beta = \frac{c}{a}\]

Answer 26

\[ax^2+bx+c = a(x-\alpha)(x-\beta) = ax^2 -a(\alpha+\beta)x + \alpha\beta \]

compare coefficients

Answer 27

As I said, this is high school maths...

Answer 28

and it's just a bit of fun

Answer 29

thanks everyone

Answer 30

this is so cool and yes highschool level :)
http://www.artofproblemsolving.com/Wiki/index.php/Newton%27s_Sums

Answer 31

tht was fun :)
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