Calculator, Q 17 - what it getting at

[thomas49th]

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

 

[cristo]

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

Pythagoras?

 

[thomas49th]

(x+8)(x+8) = x^{2} + (x+5)(x+5)

x ^ {2} + 16x + 64 = x^4+10x+25

take LHS from RHS

x^{2} - 6x - 39 = 0

but how did you know to use pythagerous?

 

[arildno]

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

 

[cristo]

(x+8)(x+8) = x^{2} + (x+5)(x+5)

x ^ {2} + 16x + 64 = x^4+10x+25

take LHS from RHS

x^{2} - 6x - 39 = 0

Sorry, one correction to your second line.

x ^ {2} + 16x + 64 = 2x^2+10x+25

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.