If $a^{2} + b^{2} + c^{2} = 35$ and $a b + b c + c a = 23$ ; find
$a + b + c$ .

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Solution

The correct option is B $\pm 9$
We know that
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
Given $a b + b c + c a = 23$
$a^{2} + b^{2} + c^{2} = 35$
$=> (a + b + c)^{2} = 35 + 2 (23)$
$=> (a + b + c)^{2} = 35 + 46$
$=> (a + b + c)^{2} = 81$
$=> (a + b + c) = \sqrt{81}$
$=> a + b + c = \pm 9$

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Find $a^{2} + b^{2} + c^{2}; i f a + b + c = 9 a n d a b + b c + c a = 24$

If a + b + c = 9 and ab + bc + ca = 23, then $a^{2} + b^{2} + c^{2}$ =

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a b + b c + c a = 26

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