If a+b+c =10 and ab+bc+ca=31 than find the value of a2+b2+c2
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Answered by
8
(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca
a^2+b^2+c^2= (a+b+c)^2-2ab-2bc-2ca
a^2+b^2+c^2= (10)^2-2(ab+bc+ca)
a^2+b^2+c^2= (10)^2-2(31)
a^2+b^2+c^2= 100-62
a^2+b^2+c^2= 38
a^2+b^2+c^2= (a+b+c)^2-2ab-2bc-2ca
a^2+b^2+c^2= (10)^2-2(ab+bc+ca)
a^2+b^2+c^2= (10)^2-2(31)
a^2+b^2+c^2= 100-62
a^2+b^2+c^2= 38
Answered by
7
Given that a+b+c=10 and ab+bc+ca=31
Then,
(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)
10^2 = a^2+b^2+c^2+2×31
100 = a^2+b^2+c^2+62
a^2+b^2+c^2 = 100-62
a^2+b^2+c^2 = 38
Then,
(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)
10^2 = a^2+b^2+c^2+2×31
100 = a^2+b^2+c^2+62
a^2+b^2+c^2 = 100-62
a^2+b^2+c^2 = 38
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