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Math Help - Help me on a question

  1. #1
    Junior Member rednest's Avatar
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    Mar 2008
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    Cool Help me on a question

    Question 5 AEA summer 2002:

    For part (a), I did -kx(2+x)=(2.5)^2-x^2, and I got (1-k)x^2-2kx-(2.5)^2=0. I then used the quadratic formula to get 4k^2-25k+25=0, giving k=5 or \frac{5}{4} (because if f touches g, then b^2=4ac). However, the next step I'm confused about. Please help. What value of k do I choose?
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello Rednest
    Quote Originally Posted by rednest View Post
    However, the next step I'm confused about. Please help. What value of k do I choose?
    The equation -kx(2+x)=2.5^2-x^2 is the same as f(x)=g(x) only if x lies in [-2,0[. If you solve the two equations -kx(2+x)=2.5^2-x^2 for x with k=5 and k=5/4 you'll get a solution x\in[-2,0[ and another one x\not\in[-2,0[ : the right value of k is the one such that x\in[-2,0[.
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