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Simple roots of a quadratic question

  1. Nov 18, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data
    Given that the roots of [itex]x^2+px+q=0[/itex] are [itex]\alpha and \beta[/itex], form an equation whose roots are [itex]\frac{1}{\alpha} and \frac{1}{\beta}[/itex]
    b) Given that [itex]\alpha[/itex] is a root of the equation [itex]x^2=2x-3[/itex] show that
    i)[itex]\alpha^3=\alpha-6[/itex]
    ii)[itex]\alpha^2-2\alpha^3=9[/itex]


    2. Relevant equations
    [tex]\sum\alpha=\frac{-b}{a}[/tex]
    [tex]\sum\alpha\beta=\frac{c}{a}[/tex]

    3. The attempt at a solution
    well for part a) I can do that simply
    but it is part b) that confuses me, as I can simply factorize what the eq'n and get 1 and -3 as the roots..but I don't see how i can use those numbers to prove what they want to show...
     
    Last edited: Nov 18, 2007
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  3. Nov 18, 2007 #2

    dynamicsolo

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    Are you quite sure the roots are 1 and -3? What is the discriminant?
     
  4. Nov 18, 2007 #3

    rock.freak667

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    ah wait...wrong sign i used...must redo

    Well the roots are imaginary, but if i take [itex]\alpha[/itex] to be one of them and cube it I wouldn't get what i want
     
    Last edited: Nov 18, 2007
  5. Nov 18, 2007 #4

    dynamicsolo

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    What exactly were the roots you found? I just tried out (b-i) and got that to work. I'm trying (b-ii) now...
     
  6. Nov 18, 2007 #5

    rock.freak667

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    1+- 2sqrt(2)i
     
  7. Nov 18, 2007 #6

    dynamicsolo

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    I find the discriminant to be -8 , so the imaginary part would be just sqrt(2).
     
  8. Nov 18, 2007 #7

    rock.freak667

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    ah...forgot to divide the imag. part by 2...but if i take [itex]\alpha[/itex] as either of the 2 roots it will work out?
     
  9. Nov 18, 2007 #8

    dynamicsolo

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    You'll find that changing the sign of the imaginary part only flips the signs of certain terms in the expressions you are evaluating in a manner that still allows them to cancel out. Try it on (b-i)...
     
  10. Nov 18, 2007 #9

    HallsofIvy

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    Of course, actually solving the equation and using those values as [itex]\alpha[/itex] defeats the point of the problem: use the equation itself!

    If [itex]\alpha[/itex] satisfies [itex]x^2= 2x- 3[/itex] then obviously [itex]\alpha^2= 2\alpha - 3[/itex] so [itex]\alpha^3= 2\alpha^2- 3\alpha[/itex]. But, again, [itex]\alpha^2= 2\alpha - 3[/itex] so [itex]\alpha^3= 2(2\alpha- 3)- 3\alpha= \alpha - 6[/itex]
     
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