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Graph to find solutions

  1. Jan 31, 2012 #1
    Hello folks, here is a problem. I don't know how and where to start solving it:
    A student is told to draw a linear graph on the same axis such that the intersection of the two graph will give the solutions to the equation x2+ 4x -7 =0. What is the equation of the linear graph he needs to draw?
    A. x=1 B. x=-1 C. y=1 D. y=-1 E. x+y=1
    I don't know where to start from. Can anyone help? Just give me hints that I will follow in proceeding towards the calculation. Please note that I mean x squared when I wrote x2. Thank you.
     
  2. jcsd
  3. Jan 31, 2012 #2

    tiny-tim

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    hello chikis! :smile:
    but what's the first graph? :confused:
     
  4. Jan 31, 2012 #3
    The first graph is a quadratic graph (y=x2+4x-6).
     
  5. Jan 31, 2012 #4

    tiny-tim

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    ok, so you need to combine y = x2 + 4x - 6 with x2 + 4x - 7 = 0 :wink:
     
  6. Jan 31, 2012 #5
    You mean I should combine them by adding it up together as in y= (x2+4x-6) + (x2+4x-7)=0. Is that what you mean?
     
  7. Jan 31, 2012 #6

    tiny-tim

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    no, i mean combine them as in simultaneous equations :smile:

    ('cos you want them to be simultaneously true :wink:)
     
  8. Jan 31, 2012 #7
    You mean I should arrange and solve them like this like if am solving simultanous equation?:
    y=x2+4x-6
    x2+4x-7=0.
    Is that what you mean?
     
  9. Jan 31, 2012 #8

    tiny-tim

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  10. Jan 31, 2012 #9
    That cannot be true because all the two equation resembles quadratic equation. In that case it (the both equation) cannot be solved simultanously. For that to be possible it means that one of the equation has to be linear and the other quadratic.
     
  11. Jan 31, 2012 #10

    tiny-tim

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    no, the second equation isn't y = x2 + 4x - 7 :wink:
     
  12. Jan 31, 2012 #11
    does that look like linear equation to you?
     
  13. Jan 31, 2012 #12

    tiny-tim

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    just solve them!!!
     
  14. Jan 31, 2012 #13
    That cannot be possible. It can't work that way. Maybe you should try it; let's see what you will get. How about that?
     
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