Register to reply

Calculator, Q 17 - what it getting at

by thomas49th
Tags: calculator
Share this thread:
thomas49th
#1
Apr7-07, 07:46 AM
P: 656

What is this question going for. I can identify that OTA is a right angled triangle..... Where do I go from
Phys.Org News Partner Science news on Phys.org
Hoverbike drone project for air transport takes off
Earlier Stone Age artifacts found in Northern Cape of South Africa
Study reveals new characteristics of complex oxide surfaces
cristo
#2
Apr7-07, 07:51 AM
Mentor
cristo's Avatar
P: 8,305
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
#3
Apr7-07, 07:55 AM
P: 656
[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
#4
Apr7-07, 07:56 AM
Sci Advisor
HW Helper
PF Gold
P: 12,016
Calculator, Q 17 - what it getting at

What relationships DO you know hold for right-angled triangles that might come in handy?
cristo
#5
Apr7-07, 07:59 AM
Mentor
cristo's Avatar
P: 8,305
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.


Register to reply

Related Discussions
TI-89 Calculator Calculators 5
Calculator help Calculators 4
Math ap Calculator Calculators 8
Do you really need a calculator? General Math 21
Exp in calculator General Math 3