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

Question

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

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6 days 2021-09-10T11:13:34+00:00 2 Answers 0 views 0

1. ➜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