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If a^2+b^2+c^2=2(a-b-c)-3 then find the value of 2a-3b+4c

If a^2+b^2+c^2=2(a-b-c)-3 then find the value of 2a-3b+4c

Grade:10

1 Answers

Arun
25750 Points
5 years ago
a^2+b^2+c^2 = 2(a-b-c)-3 
a^2+b^2+c^2-2a+2b+2c+3 = 0 
(a^2-2.a.1+1^2)+(b^2+2.b.1+1^2)+(c^2+2... = 0 
(a-1)^2+(b+1)^2+(c+1)^2 = 0 
[if the sum of three terms equals to 0, 
the squares are also equals to 0 
because square of something never be negative...] 
so, 
(a-1)^2 = 0 
.'.a = 1 
(b+1)^2 = 0 
.'.b = -1 
(c+1)^2 = 0 
.'.c = -1 

Therefore: 
2a - 3b + 4c = 2 + 3 -4 = 1

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