Quadratic

Quadratic

Posers and Puzzles

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p

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18 Jun 05

Let a and b be the roots of the quadratic equation x^2 + x - 1 = 0.
Find the value of a^6 + b^6.

M

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18 Jun 05

18

p

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18 Jun 05

Originally posted by Mephisto2
18
Thats correct. ๐Ÿ˜‰

s
Fast and Curious

slatington, pa, usa

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23 Jun 05

Originally posted by phgao
Let a and b be the roots of the quadratic equation x^2 + x - 1 = 0.
Find the value of a^6 + b^6.
isn't it minus 18? roots are +0.61803 and -1.618.

C

Earth Prime

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23 Jun 05

Originally posted by sonhouse
isn't it minus 18? roots are +0.61803 and -1.618.
no.... they are both raised to a positive exponent before adding together, therefore they will both be positive additons.

i

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23 Jun 05

Originally posted by Coconut
no.... they are both raised to a positive exponent before adding together, therefore they will both be positive additons.
I guess you mean positive "even" exponent?

s
Fast and Curious

slatington, pa, usa

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23 Jun 05

Originally posted by ilywrin
I guess you mean positive "even" exponent?
I did it on my calculator, a casio fx-115ms, and without
parenthesis it comes out as a minus. But like you said,
any minus number raised to an even exponent has to be +.
It turns out you need to put the number and minus sign inside
the parenthesis, (-1.6)^6=+17......
going -1.6^6 on the casio, no parentheisis, gives -17......

s
Fast and Curious

slatington, pa, usa

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24 Jun 05

Originally posted by phgao
Let a and b be the roots of the quadratic equation x^2 + x - 1 = 0.
Find the value of a^6 + b^6.
here is another question: can X^2+X-1 be factorized?
llike X^2-1 is (X-1)*(X+1)
x^2-2X+1 is (X-1)^2 and X^2+2X+1 is (X+1)^2
That pretty much kills all the x and 1 combinations so is there a
regular factoriztion for this formula? Already know the roots.

i

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24 Jun 05

Well IIRC any polynomial may be written as:
a(x - z1) (x-z2)...(x- zn), where a is the coefficient before the highest power of x, and z1,..,zn are all the roots.

s
Fast and Curious

slatington, pa, usa

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25 Jun 05

Originally posted by ilywrin
Well IIRC any polynomial may be written as:
a(x - z1) (x-z2)...(x- zn), where a is the coefficient before the highest power of x, and z1,..,zn are all the roots.
well, learn something new every day. so if root 1 is called R1 and
root 2 is called R2 then for x^2-X-1 it goes (X-R1)*(X-R2) right?

s
Fast and Curious

slatington, pa, usa

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25 Jun 05

funny, in all the decades of one math course after another, never
ran into that one๐Ÿ™‚

C

Earth Prime

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25 Jun 05

Originally posted by ilywrin
I guess you mean positive "even" exponent?
ummm.. yeah. Even was what I wanted.

i

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25 Jun 05

Originally posted by sonhouse
funny, in all the decades of one math course after another, never
ran into that one๐Ÿ™‚
I guess you have missed on akgebra classes. Here it is:
http://mathworld.wolfram.com/PolynomialRoots.html