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Posers and Puzzles

Posers and Puzzles

  1. 18 Jun '05 07:17
    Let a and b be the roots of the quadratic equation x^2 + x - 1 = 0.
    Find the value of a^6 + b^6.
  2. 18 Jun '05 07:41
    18
  3. 18 Jun '05 10:07
    Originally posted by Mephisto2
    18
    Thats correct.
  4. Subscriber sonhouse
    Fast and Curious
    23 Jun '05 12:29
    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.
  5. 23 Jun '05 12:51
    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.
  6. 23 Jun '05 12:59
    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?
  7. Subscriber sonhouse
    Fast and Curious
    23 Jun '05 21:51
    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......
  8. Subscriber sonhouse
    Fast and Curious
    24 Jun '05 15:37
    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.
  9. 24 Jun '05 15:42
    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.
  10. Subscriber sonhouse
    Fast and Curious
    25 Jun '05 01:40
    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?
  11. Subscriber sonhouse
    Fast and Curious
    25 Jun '05 01:42
    funny, in all the decades of one math course after another, never
    ran into that one
  12. 25 Jun '05 03:56
    Originally posted by ilywrin
    I guess you mean positive "even" exponent?
    ummm.. yeah. Even was what I wanted.
  13. 25 Jun '05 07:03
    Originally posted by sonhouse
    funny, in all the decades of one math course after another, never
    ran into that one
    I guess you have missed on akgebra classes. Here it is:
    http://mathworld.wolfram.com/PolynomialRoots.html