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# algebra

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a^2+b^2+c^2=121 and the question is: what is a+b+c=?

The answe is 17 but I don´t know how to prove it and which sort of method I should use. It should be quite easy since the course as a whole is considered to be basic

Sep 9, 2017

#1
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since a + b + c = 17, then one way could be: a=2, b=6, c=9. So, you have:

a^2 + b^2 + c^2 = 2^2 + 6^2 + 9^2 =4 + 36 + 81 =121.

Sep 9, 2017

#1
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since a + b + c = 17, then one way could be: a=2, b=6, c=9. So, you have:

a^2 + b^2 + c^2 = 2^2 + 6^2 + 9^2 =4 + 36 + 81 =121.

Guest Sep 9, 2017
#2
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Yes, I totally get that, but then is when you are aware of the answer, which is not the case from the beginning. So, how do I prove that without being aware of the asnwer?

Guest Sep 10, 2017
#3
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Without giving the condition that a+b+c=17, I don't think that you can formally "prove" it since there are other solutions that will satisfy the a^2+b^2+c^2=121. Other solutions could be a=0, b=0, c=11 and a=6, b=6 and c=7.....and so on.

Sep 10, 2017