Question

if a+b+c=5,a^+b^+c^=2a,then find the value of ab +bc+ca​

Answer 1

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➜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

thank uhh ♥️

Answer 2

Answer:

same as ponaam

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