If $a + b + c = 5$ and $a^{2} + b^{2} + c^{2} = 29$ , then the value of $a b + b c + c a$ is

- 2

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Solution

The correct option is A $- 2$
$a + b + c = 5$ $a^{2} + b^{2} + c^{2} = 29$
$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$
$25 = 29 + 2 (a b + a c + b c)$
$2 (a b + b c + a c) = - 4$
$a b + b c + a c = - 2$

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Similar questions

a + b + c = 6

a b + b c + c a = 11

a^{2} + b^{2} + c^{2}

If a = 1, b = -2 and c = 3 then find the value of a $^{2}$ +b $^{2}$ +c $^{2}$ +ab+bc+ca.

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- (\frac{a^{2} + b^{2} + c^{2}}{b c + c a + a b})

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\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

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