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Calculator, Q 17 - what it getting at

  1. Apr 7, 2007 #1
    [​IMG]
    What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
     
  2. jcsd
  3. Apr 7, 2007 #2

    cristo

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    Pythagoras?
     
  4. Apr 7, 2007 #3
    [tex](x+8)(x+8) = x^{2} + (x+5)(x+5)[/tex]

    [tex]x ^ {2} + 16x + 64 = x^4+10x+25[/tex]

    take LHS from RHS

    [tex]x^{2} - 6x - 39 = 0[/tex]

    but how did you know to use pythagerous?
     
  5. Apr 7, 2007 #4

    arildno

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    What relationships DO you know hold for right-angled triangles that might come in handy?
     
  6. Apr 7, 2007 #5

    cristo

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    Sorry, one correction to your second line.

    [tex]x ^ {2} + 16x + 64 = 2x^2+10x+25[/tex]

    Well, you did the hard part; spotting that it was a right angled triangle. Pythagoras' theorem holds for right angled triangles, and is a relationship relating the squares of the sides. Since the solution contains an x^2, this is quite a big hint as to what you should use.
     
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