If (a-1)²+(b+2)²+(c+1)²=0, then the value of 2a-3b+7c is?
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we all knew that (a-1)²,(b+2)²,(c+1)² are positive with every a,b,c
so in order to have (a-1)²+(b+2)²+(c+1)²=0, (a-1)² must = 0 ,(b+2)² must =0 and (c+1)² must =0
so a=1, b=-1, c=-1
2a-3b+ 7c = -2
so in order to have (a-1)²+(b+2)²+(c+1)²=0, (a-1)² must = 0 ,(b+2)² must =0 and (c+1)² must =0
so a=1, b=-1, c=-1
2a-3b+ 7c = -2
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