Calculator, Q 17 - what it getting at

thomas49th · Apr 7, 2007

What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from

cristo · Apr 7, 2007

thomas49th said: ↑

What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from

Pythagoras?

thomas49th · Apr 7, 2007

[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 · Apr 7, 2007

What relationships DO you know hold for right-angled triangles that might come in handy?

cristo · Apr 7, 2007

thomas49th said: ↑

[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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Calculator, Q 17 - what it getting at

thomas49th

cristo

thomas49th

arildno

cristo

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