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If a + b + c = 26 and ab + bc + ca = 109, find the value of a² + b² + c² = ?

1458
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4476

Answer:- 1

Explanation:-

Solution:

\begin{aligned} {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow {(26)}^{2} = a^{2} + b^{2} + c^{2} + 2 (109) \\ \Rightarrow a^{2} + b^{2} + c^{2} = 676 - 218 \\ \Rightarrow a^{2} + b^{2} + c^{2} = 458 \end{aligned}

$\eqalign{ & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} {(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) \\ \Rightarrow {(26)}^{2} = a^{2} + b^{2} + c^{2} + 2 (109) \\ \Rightarrow a^{2} + b^{2} + c^{2} = 676 - 218 \\ \Rightarrow a^{2} + b^{2} + c^{2} = 458 \end{aligned}

$\eqalign{ & {\left( {a + b + c} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {ab + bc + ca} \right) \cr & \Rightarrow {\left( {26} \right)^2} = {\text{ }}{a^2} + {b^2} + {c^2} + 2\left( {109} \right) \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 676 - 218 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 458 \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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