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 4

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^2-2(a-b-c)+3=0

a^2+b^2+c^2-2a+2b+2c+3=0

⇒(a^2+1-2a)+(b^2+1+2b)+(c^2+1+2c)=0

⇒(a-1)^2+(b+1)^2+(c+1)^2=0

This is possible when

(a-1)^2=0,(b+1)^2=0 \and\; (c+1)^2=0

⇒a=1,b=-1,c=-1

Thus,

2a-3b+4c=2(1)-3(-1)+4(-1)

= 2 + 3 – 4 = 1

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