If a + b + c = 12 and a^2 + b^2 + c^2 = 64 find the value of ab + bc + ac
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given
a+b+c=12 and a^2+b^2+c^2=64
a+b+c=12 (ab+bc+ac)=?
Squaring on both sides
(a+b+c)^2=144
a^2+b^2+c^2+2ab+2bc+2ac=144
substitute a^2+b^2+c^2=64 in this equation
64+2ab+2bc+2ac=144
Transpose 64 on the RHS
2ab+2bc+2ca=80
take 2 common which will be equal to
2(ab+bc+ac)=80
Therefore ab+bc+ca=40
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