If a+b+c=12 and a^{2}+b^{2}+c^{2}=100 find the value of ab+bc+ca
Answers
Step-by-step explanation:
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Answer:
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Step-by-step explanation:
Here, we having two equations
a + b + c = 12 ...(eq.1)
a^2 + b^2 + c^ =64 …(eq.2)
by eq.1 we can calculate the value of a, b & c as
a = 12 - b - c
b = 12 - a - c
c = 12 - a - b
then we put the value of a in second equation.
we get,
(12 - b - c)^2 + b^2 + c^2 = 64
after solving this,
we get, b^2 + bc + c^2 - 12b - 12c + 40 =0
by using above equation we calculate the term bc.
therefore, bc = 12b + 12c - b^2 -c^2 - 40
similarly by putting the values of b and c in eq.2 we get,
ac = 12a + 12c - a^2 - c^2 - 40
ab = 12a + 12b - a^2 - b^2 - 40
and now,
ab + bc + ac = (12a + 12b - a^2 - b^2 - 40) + (12b + 12c - b^2 -c^2 - 40) + (12a + 12c - a^2 - c^2 - 40)
ab + bc + ac = 24a + 24b + 24c - 2a^2 - 2b^2 – 2c^2 – 120
ab + bc + ac = 24( a + b + c ) - 2(a^2 + b^2 + c^) - 120
by eq.1 and eq.2,
ab + bc + ac = 24(12) - 2(64) - 120
ab + bc + ac = 40
Hence, ab + bc + ac = 40