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Simulteneous equation

  1. Aug 12, 2010 #1
    Show that the simultaneous equation always have two distinct solutions, for all possible values of k

    2x^2+ xy = 10

    x + y = k

    I know that i have to use b^2 - 4ac = 0

    please try the number i am having some problems...
    my work

    y = k - x

    so i replace in equation and i got:

    2x^2 + xk - x^2 - 10 = 0 as from here i am stuck..help how to prove that..????
     
  2. jcsd
  3. Aug 12, 2010 #2

    thrill3rnit3

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    Gold Member

    2x2 + kx - x2 - 10 = 0

    simplifies to

    x2 - kx - 10 = 0

    which is a quadratic equation for any k

    the quadratic formula has the +/- thing, so you either have 2 real roots or 2 complex roots [ depending on the value of the discriminant ]
     
  4. Aug 13, 2010 #3

    hey how do you obtain -kx
     
  5. Aug 13, 2010 #4

    thrill3rnit3

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    I meant +kx sorry...

    But the rest is still the same.
     
  6. Aug 15, 2010 #5
    Yeahh me too i have reacheed this level...but the problem is that when i used the formula b^2 - 4ac i cannot get the answer..!!!!!!


    a = -1

    b = -k

    c = 10

    (-k)^2 - 4 (-1)(10) = 0

    k^2 = -40

    so i am stuck here...what should i do
     
  7. Aug 15, 2010 #6

    Borek

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    Staff: Mentor

    You don't solve for k, you solve for x,y - and you don't give answer as a number (you can't not knowing value of k parameter), but as a formula containing k.
     
  8. Aug 15, 2010 #7

    HallsofIvy

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    [itex]b^2- 4ac= k^2+ 40[/itex] which is always positive.
     
  9. Aug 15, 2010 #8
    You do not have to solve for any value. All you have to do is show that there are two distinct solutions for any value of k. This means you must show that discriminant

    [tex]D = b^2 - 4 a c = k^2 + 40 > 0 [/tex] for all values of k. Is this true?
     
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