If $a^{2} + b^{2} + c^{2} = 16$ and ab + bc + ca = 10, find the value of a + b + c.

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Solution

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 16 + 2 \times 10 = 16 + 20 = 36 = (\pm 6)^{2} ∴ a + b + c = \pm 6$

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