1f (a+b-6)2 + a2+b2+1+2b = 2ab+2a, then the value of a is
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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.,
Value of a=6-b and a=b+1
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