If $a^2+b^2+49 c^2+18=2(b+28 c-a)$, then the value of $(2 a-b+7 c)$ is:
Answer & explanation
Correct answer: option 4
If $a^2+b^2+49 c^2+18=2(b+28 c-a)$
$(2 a-b+7 c)$ = ?
a2 + b2 + 49c2 + 18 = 2(b - 28c - a)
= a2 + b2 + 49c2 + 18 = 2b - 56c -2a
= a2 + b2 + 49c2 + 18 - 2b + 56c + 2a = 0
= (a2 + 2a + 1 ) + (b2 - 2b + 1) + (7c)2 + 56c + 16 = 0
= (a + 1)2 + (b - 1)2 + {(7c)2 + 56c + 42} = 0
= (a + 1)2 + (b - 1)2 + (7c + 4)2 = 0
Now,
a + 1 = 0
= a = -1
b -1 = 0
= b = 1
7c + 4 = 0
= c = \(\frac{4}{7}\)
$(2 a-b+7 c)$ = (2 × -1 - 1 +7 × \(\frac{4}{7}\)) = 1