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If a^2+b^2+c^2=2(a+2b-2c)-9,then what is the value of a+b+c?
a^2 + b^2 + c^2 – 2a – 4b +4c + 9 = 0{By opening the brackets and bringing the terms to one side}Now you can see that we may break 9 = 4 + 4 + 1 (In order to make whole sqaures of a , b , c )We have(a² -2b + 1) + (b²– 4b +4) + (c² + 4c +4 ) =0(a-1)² + (b-2)²+ (c+2)²= 0Now we know (a-1)² ,(b-2)² ,(c+2)² are greater than or equal to 0 for all real values of a,b,c.So to make this whole Equation zero these individual squares will have to be 0 .From here we get a = 1; b= 2 and c= -2 and a+b+c = 1Tell me if you got this!!
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