Question

215 If a+b+c=6 and ab+bc+ca = 11, find a'^2+b^2+c^2
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Answer 1

Answer :D

Given: a + b + c = 6, ab + bc + ca = 11

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

6² = a² + b² + c² + 2(11)

36 = a² + b² + c² + 22

a² + b² + c² = 14

Answer 2

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GIVEN,

$= > a + b + c = 6 \\ = > ab + \: bc + \: ca = 11$

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TO FIND,

$= > a ^{2} + b ^{2} + c ^{2}$

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IDENTITY USED,

$= > (a + b + c)^{2} = a ^{2} + b ^{2} + c ^{2} + 2ab + 2bc + 2ca$

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SOLUTION,

$= > a ^{2} + b ^{2} + c ^{2} + 2ab + 2bc + 2ca = (a + b + c) ^{2} \\ \\ = > a ^{2} + b ^{2} + c ^{2} + 2(ab + bc + ca) = (6) ^{2} \\ \\ = > a ^{2} + b ^{2} + c ^{2} + 2 \times 11 = 36 \\ \\ = > a ^{2} + b ^{2} + c ^{2} + 22 = 36 \\ \\ = > a ^{2} + b ^{2} + c ^{2} = 36 - 22 \\ \\ = > a ^{2} + b ^{2} + c ^{2} = 14$

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