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Question

If a,b and c are real number such that a2+b2+2c2=4a-2c+2bc-5,then find the possible value of (6a-4b-2c).

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Solution

a² + b² + 2c² = 4a -2c + 2bc - 5
a² + b² + 2c² – 4a + 2c – 2bc + 5 =0
complete the square
(a² – 4a + 4) + (b² – 2bc + c²)– 2c² + 2c + 5 – 4 – c² =0
(a – 2)² + (b – c)² + (c + 1)²=0.

Since each term is a square each is nonnegative. Since their sum is zero each must be zero. So a = 2 and b = c = -1.
(6a-4b-2c).=(6*2)-(4*-1)-(-2*-1)
=12+4-2
=14

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