Question

if a+b+c=13 and ab+bc+ca=84 find a^2+b^2+c^2​

Answer 1

Answer:

given a+b+c = 13 and ab+bc+ca = 84

to find a²+b²+c²

(a+b+c)² = a²+b²+c² + 2(ab+bc+ca)

13² = a²+b²+c² + 2(84)

169 = a²+b²+c²+168

therefore a²+b²+c² = 169 - 168 = 1

Answer 2

Answer:

1

Step-by-step explanation:

$Given, \\ a + b + c = 13 \\ ab + bc + ca = 84 \\ Since, {(a + b + c )}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ca ) \\ = > {13}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2 \times 84 \\ = > 169 = {a}^{2} + {b}^{2} + {c}^{2} + 168 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 169 - 168 \\ = > {a}^{2} + {b}^{2} + {c}^{2} = 1$

Hope it helps you......✔✔✔

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