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Question 124
If $$a^2+b^2+4c^2=2(a+b-2c-3)$$Â and a, b, c are real, then the value of $$(a^2+b^2+c^2)$$Â is
Solution
$$a^2+b^2+4c^2=2(a+b-2c-3)$$
So, $$(a-1)^2+(b-1)^2+(2c+1)^2=0$$
Or, $$a=1,b=1,c=\frac{-1}{2}$$
So, $$a^2+b^2+c^2=1+1+\frac{1}{4}=\frac{9}{4}$$
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