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Systems of Quadratics

  1. Jan 27, 2005 #1
    :eek: OMG! This question is annoying me to death... :bugeye:

    x^2 + y^2 = 25
    x^2 + y = -5

    I understand you substitute, and I did...
    x^2 = -5 - y

    So I plugged it in...
    (-5 - y) + y^2 = 25
    subtract 25) y^2 - y - 30 = 0

    I tried the quadratic formula AND did the whole (y + 1)(y - 30)=0,
    but they would not equal the answers: (3,4)(-3,4)(-5,0)

    Please help! I can't understand why I can't get this answer, but yet I can get the others. :surprised:
     
  2. jcsd
  3. Jan 27, 2005 #2
    I tried the quadratic formula AND did the whole (y + 1)(y - 30)=0,

    wrong...
    (y+1)(y-30)= y^2 -29y -30
    do it again pls
     
  4. Jan 27, 2005 #3
    okay....i see...
    (y-6)(y+5)=0
    So I plugged in:
    6) x^2 = -5 - (6)
    x^2 = -11
    I also plugged in -5) x ^2 = -5 - (-5) = 0
    That got me the (-5,0), but I still do not understand how to get the other two points...
     
  5. Jan 27, 2005 #4
    see if you copy your question correctly,
    x^2 + y^2 = 25
    x^2 + y = -5
    has one solution only , which is (0,-5), if you don't believe me, you can try plug in your ANSWER in the equations above, you will see LHS is NOT equal RHS
     
  6. Jan 27, 2005 #5
    ack....

    :yuck: yea..i copied the question wrong...it was:
    x^2 + y ^2 = 25
    y - x^2 = -5
    Got the answers, thnx SO much!

    I STILL can't believe I copied the wrong question...sorry I put you through the trouble with that, too...
     
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