Question

a+b+c=9,ab+bc+ca=40 find a^2+b^2+c^2

Answer 1

$a + b + c = 9 \\ ab + bc + ca = 40 \\ (a + b + c) {}^{2} = 9 {}^{2} \\ a {}^{2} + b {}^{2} + c{}^{2} + 2(ab + bc + ca) = 81 \\ a {}^{2} + b {}^{2} + c {}^{2} + 2 \times 40 = 81 \\ a {}^{2} + b {}^{2} + c {}^{2} + 80 = 81 \\ a { }^{2} + b {}^{2} + c {}^{2} = 81 - 80 \\ a {}^{2} + b {}^{2} + c {}^{2} = 1$

Answer 2

Hello Dear!!!

Here's your answer...

Given that

a+b+c = 9

ab + bc + ca = 40

We know that,

(a+b+c)² = a² + b² + c² + 2(ab+bc+ca)

(9)² = a² + b²+c²+2(40)

81= a² + b² + c² + 80

81-80 = a² + b² + c²

a² + b² + c² = 1

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HOPE THIS HELPS YOU....