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Q.

If $(a + b + c) = 8$ and $(a b + b c + c a) = 19,$ then $(a^{2} + b^{2} + c^{2})$ is [[1]].

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answer is 26.

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Detailed Solution

detailed_solution_thumbnail

If

(a + b + c) = 8

and

(a b + b c + c a) = 19,

then

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

is 26.
It is given that

(a + b + c) = 8

and

(a b + b c + c a) = 19.

We know that

{(p + q + r)}^{2} = p^{2} + q^{2} + r^{2} + 2 (p q + q r + p r) .

Considering here

p = a, q = b a n d r = c

and then substituting in the given formula, we get,

Therefore on substitution, we get,

\Rightarrow a^{2} + b^{2} + c^{2} = 26

Therefore, the value of

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

is 26.

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If a+b+c =8 and ab+bc+ca =19, then a 2 + b 2 + c 2 is [[1]].