Question

If (a+b-6)^2 + a^2 +b^2 +1+2b = 2ab+2a, then the value of a is

Answer 1

Given (a+b-6)² + a² +b² +1+2b = 2ab+2a

=> (a+b-6)² + a² +b² +1+2b -2ab-2a = 0

=> (a+b-6)² + [(-a)² +b² +1²+2× (-a)×b+2×b×1+2×1×(-a )]= 0

=> (a+b-6)² + (-a+b+1)² = 0

/* If Sum of the squares are equal zero then each term is equal to zero */

a + b - 6 = 0 and -a + b + 1 = 0

=> a = 6 - b and a = b + 1

Therefore.,

$\green {Value \:of \: a = 6 - b \:and \:a = b + 1 }$

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