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Does not (17+r)^2 equal 289 +r^2?
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So the problem is r^2 + 24^2 = (17 + r)^2
My teacher says it's is wrong.
No, \[(a+b)² = a² + 2ab + b²\]. So, \[(17 + r)² = 17² + 2*17*r + r² = 279 + 34r + r²\]
It's not 279 in my answer, its 289, sorry...
whoops yeah i typed it wrong
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but wouldn't (r)^2 equal r^2?
It depends. For example: \[(-3)² = 9\]\[-3² = -9\]
where did the 2 x 17 x r come from?
oh
Like i said, \[(a+b)² = a² + 2ab + b²\] Changing a to 17 and b to r, you wil have:\[(17 + r)² = 17² + 2*17*r + r²\]
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oh... i guess that makes sense...
You can see doing: \[(17 +r)²=(17+r)(17+r)=17*17+17r+17r+r² = 289 + 34r +r²\]
ah! got it now! thanks!
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