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mathmath333

  • one year ago

quadratic equation

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& \normalsize \text{If}\ p\ \text{and}\ q\ \text{are the roots of the equation} \hspace{.33em}\\~\\ & x^2+px+q=0,\normalsize \text{then find the value of}\ p.\ \hspace{.33em}\\~\\ \end{align}}\)

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}& p+q=-p\hspace{.33em}\\~\\ & pq=p \hspace{.33em}\\~\\ \end{align}}\) from the sum and product of roots i got this , is this correct.

  3. anonymous
    • one year ago
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    yes surely...

  4. anonymous
    • one year ago
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    For equation : \(ax + by + c = 0\), Sum of Roots = \(\frac{-b}{a}\) Product of Roots = \(\frac{c}{a}\)

  5. anonymous
    • one year ago
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    Are you sure p = 1??

  6. mathmath333
    • one year ago
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    \(p=-1/2\)

  7. anonymous
    • one year ago
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    If p = 1 then from equation 2, q = 1.. But from equation 1, 2p = -q does not get satisfied.

  8. anonymous
    • one year ago
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    This looks good.. :)

  9. mathmath333
    • one year ago
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    lol but the answer given is wrong by book it says the answer is \(\large \color{black}{\begin{align}p=\{0,1,-\dfrac{-1}{2}\} \hspace{.33em}\\~\\ \end{align}}\)

  10. anonymous
    • one year ago
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    0 and -1/2 are all right, but how p = 1 is possible?

  11. mathmath333
    • one year ago
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    how did u find \(0\) as a solution.

  12. anonymous
    • one year ago
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    Wait, there is a slight mistake that you have done..

  13. anonymous
    • one year ago
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    And I also being mad, I did not check that.. What you got for Product of Roots?

  14. anonymous
    • one year ago
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    \[pq = q\] This must be the product no??

  15. mathmath333
    • one year ago
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    ?

  16. anonymous
    • one year ago
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    See, the formulae I gave above, product of roots = c/a.. in your question c = q and a = 1..

  17. Empty
    • one year ago
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    The only way \(pq=p\) can be true is if: \(q=1\) OR \(p=0\)

  18. mathmath333
    • one year ago
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    ok how about \(p=1\)

  19. anonymous
    • one year ago
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    @Empty that should be q on right side, not p..

  20. mathmath333
    • one year ago
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    oh ur right .

  21. Empty
    • one year ago
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    ? \(pq=p\) is the same thing as \(p=qp\)

  22. anonymous
    • one year ago
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    Just solve these equations: \[p + q = -p \\ pq = q\]

  23. anonymous
    • one year ago
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    How they are same?

  24. Empty
    • one year ago
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    Ohh I see, I switched out my p's and q's. Oh well haha, same idea really since they're completely symmetric. XD

  25. anonymous
    • one year ago
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    @mathmath333 if you use pq = p (the original one you did above), p = 1 as solution is not possible in any case..

  26. mathmath333
    • one year ago
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    ok i got \(p=0,1\) how about \(p=-1/2\) now ?

  27. anonymous
    • one year ago
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    That should be pq = q...

  28. Empty
    • one year ago
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    \(\large \color{black}{\begin{align}& p+q=-p\hspace{.33em}\\~\\ & pq=p \hspace{.33em}\\~\\ \end{align}}\) What mathmath wrote is wrong here, I went off this, but really the last equation is what @waterineyes is trying to tell us!

  29. mathmath333
    • one year ago
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    lol big drama

  30. mathmath333
    • one year ago
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    why \(p=1\) is not possible ?

  31. Empty
    • one year ago
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    p=1 is possible only when q=0. So you have to work out the consequence of that in the other equation.

  32. mathmath333
    • one year ago
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    its so confusin

  33. anonymous
    • one year ago
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    Hey @mathmath333 listen to me, what is product of roots you get?

  34. mathmath333
    • one year ago
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    \(pq=q\)

  35. anonymous
    • one year ago
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    yea that is great..

  36. anonymous
    • one year ago
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    Now, you get the cases of 0 and 1 ??

  37. mathmath333
    • one year ago
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    yes i do

  38. anonymous
    • one year ago
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    So, we left with how p = -/12 right?

  39. mathmath333
    • one year ago
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    yes

  40. anonymous
    • one year ago
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    *-1/2..

  41. anonymous
    • one year ago
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    Put p = -1/2 in first equation and find q...

  42. mathmath333
    • one year ago
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    \(q=1\)

  43. anonymous
    • one year ago
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    Good...

  44. anonymous
    • one year ago
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    But it is not satisfying the other equation.. :P

  45. mathmath333
    • one year ago
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    how ?

  46. anonymous
    • one year ago
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    What you get when you do p times q ?? -1/2 times 1 = -1/2 but on other side you have q which is 1 and -1/2 not equal to 1..

  47. anonymous
    • one year ago
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    \[pq = q \\ \frac{-1}{2} \times 1 = 1 \\ \frac{-1}{2} \ne 1\]

  48. anonymous
    • one year ago
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    @Empty if you have something to correct or say, then you can proceed, I have tried my bit. :)

  49. mathmath333
    • one year ago
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    wait

  50. anonymous
    • one year ago
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    Yeah waiting, ordering coffee or tea something? :P

  51. mathmath333
    • one year ago
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    is \(0\) and \(1\) both correct

  52. anonymous
    • one year ago
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    @SolomonZelman your help is needed here..

  53. mathmath333
    • one year ago
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    he came first but left

  54. anonymous
    • one year ago
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    Yes 0 and 1 both are satisfying but -1/2 is not..

  55. mathmath333
    • one year ago
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    seee this wolfram link http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

  56. mathmath333
    • one year ago
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    @waterineyes

  57. anonymous
    • one year ago
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    Now wolfram will tell the secrets of us that we don't know anything. :P

  58. mathmath333
    • one year ago
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    my head hurts

  59. mathmath333
    • one year ago
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    @ganeshie8

  60. anonymous
    • one year ago
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    And I call the first master : Please save us : @mukushla ..

  61. freckles
    • one year ago
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    \[p+q=-p \\ pq=q \] these are the equations agreed?

  62. mathmath333
    • one year ago
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    yes

  63. freckles
    • one year ago
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    \[\text{ the second equation we have } \\ pq=q \\ pq-q=0 \\ q(p-1) =0 \implies q=0 \text{ or } p=1 \\ \text{ going back to first equation } \\ \text{ if } q=0 \text{ we have } p=-p \text{ which forces } p=0 \\ \text{ if } p=1 \text{ then } 1+q=-1 \implies q=-2 \\ \text{ so far we have the following pairs of solutions } \\ (p,q) \\ (0,0) \\ (1,-2)\]

  64. freckles
    • one year ago
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    I think though you say we have another solution

  65. freckles
    • one year ago
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    so let's see if we can play with the equations more to find it

  66. mathmath333
    • one year ago
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    i thhink that is same as of frekles

  67. freckles
    • one year ago
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    \[p=\frac{-1}{2} \text{ trying first } \\ \frac{-1}{2}+q=-(\frac{-1}{2}) \\ \frac{-1}{2}+q=\frac{1}{2} \\ q=1 \\\text{ so if } p=\frac{-1}{2} \text{ is a solution } \\ \text{ then we should get } q=1 \text{ for the second equation too} \\ \frac{-1}{2}q=q \\ \frac{-1}{2}q-q=0 \\ q=0 \\ \text{ so } p=\frac{-1}{2} \text{ doesn't seem to work out }\]

  68. freckles
    • one year ago
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    the only solutions I think there to be is (0,0) and (1,-2) where those ordered pairs are in the form (p,q)

  69. mathmath333
    • one year ago
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    but wolfram and my book does say \(p=-1/2\) also.

  70. freckles
    • one year ago
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    can i see the link to wolfram

  71. freckles
    • one year ago
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    because my wolfram checks my solutions just fine

  72. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=p%2Bq%3D-p+%2C+pq%3Dq

  73. mathmath333
    • one year ago
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    http://www.wolframalpha.com/input/?i=solve+2p%5E2%2Bq%3D0%2Cq%5E2%2Bpq%2Bq%3D0+over+reals

  74. freckles
    • one year ago
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    ok I see what you did: you pluggged p and q into the quadratic and didn't use veita's formula to find p,q for the -1/2,-1/2 solution hmm... so let's see you have p=-1/2 and q=-1/2 let's see if I can solve that other system by hand ..

  75. anonymous
    • one year ago
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    If you use both p and q values, then also I think both equations will not get satisfied..

  76. mathmath333
    • one year ago
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    but vieta's formula shouls work too, something is missing

  77. freckles
    • one year ago
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    \[2p^2+q=0 \\ q^2+pq+q=0 \\ 2p^2=-q \\ -2p^2=q \\ (-2p^2)^2+p(-2p^2)+(-2p^2)=0 \\ 4p^4-2p^3-2p^2=0 \\ 2p^2(2p^2-p-1)=0 \\ p=0 \text{ or } 2p^2-p-1=0 \\ p=0 \text{ or } (2p+1)(p-1)=0 \\ p=0 \text{ or } p=\frac{-1}{2} \text{ or } p=1 \] hmm... I don't know why vieta's formula isn't working for that one p

  78. mathmath333
    • one year ago
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    :D

  79. freckles
    • one year ago
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    but I would like to know why it isn't working for veita's formula :(

  80. mathmath333
    • one year ago
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    thats a sign of good mathmatician.

  81. freckles
    • one year ago
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    http://www.wolframalpha.com/input/?i=x%5E2-1%2F2x-1%2F2%3D0 since p=-1/2 then q=-2p^2=-2(-1/2)^2=-2(-1/4)=-1/2 so we have p=-1/2 and q=-1/2 but that wolfram link doesn't give the roots -1/2 and -1/2 if gives the roots -1/2 and 1

  82. freckles
    • one year ago
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    so I think p=-1/2 shouldn't be a solution

  83. mathmath333
    • one year ago
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    :(

  84. freckles
    • one year ago
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    so veita's formula I think does work and the other way gave too many solutions (solutions that needed to be checked)

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