Question

If (\(ab+bc+ca)\) = 26 and \(\ a^2+b^2+c^2\) = 29,
then (\(a+b+c\)) is :

Answer 1

Given, (\(ab+bc+ca\)) = 26 and \(\ a^2+b^2+c^2\) = 29.
Using the identity \((\ a+b+c)^2\) = \(\ a^2+b^2+c^2+2ab+2bc+2ca\),
\(\Rightarrow\) \((\ a+b+c)^2\) = 29 + 2\(\times\)26
\(\Rightarrow\) \((\ a+b+c)^2\) = 81
\(\Rightarrow\) (\(a+b+c\)) = \(\sqrt{81}\) = 9.