Question

a2 + b2+ c2=16, ab+bc+ac=10 ,a+b+c=

Answer 1

$a {}^{2} + {b}^{2} + {c}^{2} = 16 \\ ab + bc + ca = 10 \\ (a +b + c) {}^{2} = {a}^{2} + {b}^{2} + {c}^{2} \\ + 2( ab+ bc+ca ) \\ = 16 + 2(10) \\ = 16 + 20 = 36 \\ ( a+ b+ c) {}^{2} = 36 \\ a + b + c = \sqrt{36} \\ a + b + c = 6$

Answer 2

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

= 16 + 2(ab+bc+ca)

= 16 + 2(10)

= 16 + 20

= 36

Therefore , (a+b+c)2 = 36

root of both the sides

a+b+c = 6

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