Question

a+b+c=9 and ab+bc+ca=40 then find the value of a2+b2+c2

Answer 1

a + b + c = 9

Square on both sides,

( a + b + c )^2 = 9^2

⇒ a^2 + b^2 + c^2 + 2 ( ab + bc + ca ) = 81

⇒ a^2 + b^2 + c^2 + 2( 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

We know that
(a+b+c ) ^2=a ^2+b^2+c^2+2ab+2bc+2ca
(a+b+c)^2=a^2+b^2+c^2+2 (ab +bc+ca)
(9)^2=a^2+b^2+c^2+2×40
81=a^2+b^2+c^2+80
81-80=a^2+b^2+c^2
1=a^2+b^2+c^2.
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