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Eliminating y

  1. Apr 7, 2007 #1
    [​IMG]

    For (a) I got x = [tex]\sqrt{-11}[/tex] which you cant do as you cant square root a negitive number, I cant help but feel im wrong
     
  2. jcsd
  3. Apr 7, 2007 #2

    cristo

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    Well, for that reason I would say that Bill is wrong; not you!
     
  4. Apr 7, 2007 #3
    and for (b) I began be substituting y =2x - 2 into the y²

    x² + (2x-2)(2x-2) = 25

    5x² -8x -21 = 0

    am I doing the right thing so far. When I used the quadratic formula I got a decimal number, not a whole number and the question doesn't say anything about rounding to a degree of accuracy so I presume the answer is whole numbers :eek:
     
  5. Apr 7, 2007 #4

    arildno

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    No, you may presume that the answer is to be given exactly in b)!

    That is you have the two solutions:
    [tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22}}{10}[/tex]
    As it happens, you get rational solutions here, otherwise, the exact solutions would involve square root symbols explicitly.
     
  6. Apr 7, 2007 #5

    cristo

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    What do you get when you solve this quadratic equation? The answer does not necessarily have to be a whole number, especially if it is only a short decimal. The solution to this equation is such that it can easily be written exactly.
     
  7. Apr 7, 2007 #6

    HallsofIvy

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    Or, to put it another way,
    5x2- 8x+ 21= (5x+ 7)(x- 3)= 0
     
  8. Apr 7, 2007 #7
    i must of made an error in typing it into my calculator. For one of x's solutions I got somthing like 4.926537173 (i pressed random keys after the 3.s.f). Ill check over my work now...
     
  9. Apr 7, 2007 #8
    okay, when typing it in I think I calculated the stuff inside the root wrong

    anyway the answer:

    x = -7 or 3
    y = -16 or 4
     
  10. Apr 7, 2007 #9

    arildno

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    Eeeh??
    Whatever are you talking about?
     
  11. Apr 7, 2007 #10
    never mind, but ive posted the (i think correct answers now above).

    However im concerned that -16² + -7² dont equal 25
     
  12. Apr 7, 2007 #11

    HallsofIvy

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    arildno suggested
    [tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22 }}{10}[/tex]

    and I told you that 5x2- 8x+ 21= (5x+ 7)(x- 3)= 0.

    How could you possibly get "x = -7 or 3" from that?
     
  13. Apr 7, 2007 #12
    (8 + 22)/10 = 3
    (8-22)/10 = 1.4

    woops

    EDIT: Making y 4 or 4.8
     
  14. Apr 7, 2007 #13

    HallsofIvy

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    Well, I would have said 7/5 but but I grew up BC (before calculators).
     
  15. Apr 7, 2007 #14

    arildno

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    Actually, I would say -7/5..
     
  16. Apr 8, 2007 #15
    [tex]x=\frac{-(-8)\pm\sqrt{(-8)^{2}-4*5*(-21)}}{2*5}=\frac{8\pm\sqrt{484}}{10}=\frac{8\pm{22 }}{10}[/tex]

    8+22 = 30/10 = 3
    8-22 = -14/10 = -1.4

    feed that into the equation y = 2(x) - 2

    2(3) - 2 = 6- 2 = 4
    2(-1.4) - 2 = -2.8-2 = -4.8

    I would think that must be right?
     
  17. Apr 8, 2007 #16

    HallsofIvy

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    Yes, that is correct. Now be sure to pair them correctly: the solutions to the pair of equations is x= 3, y= 4 and x= -1.4, y= -4.8.
     
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