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Find the value of k in the equation

  1. Dec 24, 2013 #1
    1. The problem statement, all variables and given/known data

    If the quadratic equations has 2 roots one of which is the square of the other find the value of k

    2. Relevant equations


    x^2-x+k=0

    3. The attempt at a solution

    let p and p^2 be the roots..

    sum of roots= p(p+1)=1

    product= p^3=k

    then solving for p[quadratic] and then substituting into p^3=k

    but the cubing part of p is rather tedious and not yielding the desired answer!

    Where am i wrong
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 24, 2013 #2
    Are you familiar with these relationships?;

    α + β = -b/a

    αβ = c/a

    Where α and β are the roots of the quadratic and a, b, c correspond to the coefficients (ax^2 + bx + c)
     
  4. Dec 24, 2013 #3

    haruspex

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    That's what was used in the OP.
     
  5. Dec 24, 2013 #4

    haruspex

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    Please post your answer and the given answer.
    There is a better way than cubing p. Use the two equations you have to eliminate the higher powers of p successively.
     
  6. Dec 24, 2013 #5
    Right you are.

    I missed that on account of both roots using p.
     
  7. Dec 24, 2013 #6
    hmm okay a better way.

    p^3=k

    p^3-1=k-1

    (p-1)(p^2+1+p)=k-1


    (-P^2)(-2)=k-1


    so that reduces p^3 to p^2 alright :D

    have i cracked the quest for the simpler way???
     
  8. Dec 24, 2013 #7

    HallsofIvy

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    It seems to me that simplest is
    [tex](x- p)(x- p^2)= x^2- x+ k[/tex]
    [tex]x^2- (p+ p^2)x+ p^3= x^2- x+ k[/tex]

    So [itex]k= p^3[/itex] and [itex]p+ p^2= 1[/itex]

    Use [itex]p+ p^2= 1[/itex] to determine p and then [itex]k= p^3[/itex] to determine k.
     
  9. Dec 24, 2013 #8
    hey did we not get that directly from sum of roots and product of roots.

    the question was whether you could avoid doing the cubing part..

    thanksss
     
  10. Dec 24, 2013 #9
    Hi Shivam!

    You have ##p^3=k##. Rewrite it as ##p^3-1=k-1##. Does that give you a hint? ;)

    EDIT: Okay, I see you have done this.
    Good so far but I don't understand what you did next. What is the value of ##p^2+p+1##?
     
  11. Dec 24, 2013 #10


    write that as -2 :D
     
  12. Dec 24, 2013 #11
    It is not -2, check again!
     
  13. Dec 24, 2013 #12
    Okay, +2 then!
     
  14. Dec 24, 2013 #13
    Yes, so have you solved the problem?
     
  15. Dec 24, 2013 #14
    yes that's why i stopped commenting
    ;p
     
  16. Dec 24, 2013 #15
    yes that's why i stopped commenting
    ;p
     
  17. Dec 24, 2013 #16

    haruspex

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    Fwiw, what I had in mind was:
    ##p^2=1-p##
    ##p^3=p-p^2=p - (1-p)##
    ##k = p^3 = 2p-1##
     
  18. Dec 25, 2013 #17
    so in the end there are two legitimate values of k, is that correct ?
     
  19. Dec 25, 2013 #18

    Haa this looks just like the reducing in matrices determints. using properties.
    Will surely keep this "trick" in mind
     
  20. Dec 26, 2013 #19

    ehild

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    You can also start with the quadratic formula.

    [tex]x_{1,2}=\frac{1\pm\sqrt{1-4k}}{2}[/tex]

    x1=x22---> [tex]\frac{1+\sqrt{1-4k}}{2}=\left(\frac{1-\sqrt{1-4k}}{2}\right)^2[/tex]
    Expanding, simplifying leads to [itex]\sqrt{1-4k}=-k[/itex], that means k<0.

    k=-2-√5.
    The roots are 0.5(√5+3 )and -0.5(√5+1).

    ehild
     
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