Discussion
The sum of the squares of three numbers is 138, while the sum of their products taken two at a time is 131. Their sum is:
Answer: B
Let the numbers be \(a\), \(b\) and \(c\).
Then,
\(a^2 + b^2 + c^2 = 138\\
\text{and}\\
ab+bc+ca = 138\\
(a + b + c)^2\\
= a^2 + b^2 + c^2 + 2(ab + bc + ca)\\
=138 + 2 \times 131\\
= 400\\
\therefore (a + b + c) = \sqrt{400} = 20\)
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