If a2 + b2 + c2 = 2(a - b - c) -3, then the value of a + b + c is?
A. -2
B. 1
C. 2
D. -1
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {a^2} + {b^2} + {c^2} = 2\left( {a - b - c} \right) - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} = 2a - 2b - 2c - 3 \cr & \Rightarrow {a^2} + {b^2} + {c^2} - 2a + 2b + 2c + 3 = 0 \cr & \Rightarrow {a^2} - 2a + 1 + {b^2} + 2b + 1 + {c^2} + 2c + 1 = 0 \cr & \Rightarrow {\left( {a - 1} \right)^2} + {\left( {b + 1} \right)^2} + {\left( {c + 1} \right)^2} = 0 \cr & {\left( {a - 1} \right)^2} = 0{\text{ }}{\left( {b + 1} \right)^2} = 0{\text{ }}{\left( {c + 1} \right)^2} = 0 \cr & \Rightarrow a - 1 = 0{\text{ }} \Rightarrow b + 1 = 0{\text{ }} \Rightarrow c + 1 = 0 \cr & \Rightarrow a = 1{\text{ }} \Rightarrow b = - 1{\text{ }} \Rightarrow c = - 1 \cr & \therefore a + b + c \cr & = 1 - 1 - 1 \cr & = - 1 \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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