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

RRB Junior Engineer Stage 1 Model Papers 2

Question : 16 of 99

Marks: +1, -0

a^2 + b^2 + c^2 = 2 (a – b – c) – 3

then the value of

2a – 3b + 4c

Solution:

a^2 + b^2 + c^2 = 2 (a – b – c) – 3

⇒ a^2+ b^2+ c^2= 2a + 2b + 2c + 3 = 0

⇒ a^2– 2a + 1 + b^2+ 2b + 1 + c^2+ 2c+ 1 = 0

⇒ (a –1)^2+ (b + 1)^2+ (c + 1)^2= 0

[ If

x^2+ y^2+ z^2= 0 ⇒ x = 0; y = 0; z = 0

]

∴ a – 1 = 0 ⇒ a = 1

b + 1 = 0 ⇒ b = – 1

c + 1 = 0 ⇒ c = – 1

∴ 2a – 3b + 4c = 2 + 3 – 4 = 1

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