Calculator, Q 17 - what it getting at


by thomas49th
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thomas49th
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Apr7-07, 07:46 AM
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What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
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cristo
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Apr7-07, 07:51 AM
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Quote Quote by thomas49th View Post
What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
Pythagoras?
thomas49th
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Apr7-07, 07:55 AM
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[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?

arildno
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Apr7-07, 07:56 AM
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Calculator, Q 17 - what it getting at


What relationships DO you know hold for right-angled triangles that might come in handy?
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Apr7-07, 07:59 AM
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Quote Quote by thomas49th View Post
[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]
Sorry, one correction to your second line.

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

but how did you know to use pythagerous?
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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