If $$\frac{{{a^2} + {b^2} + {c^2} - 1024}}{{ab - bc - ca}} = - 2$$ and a + b = 5c, where c > 0, then the value of c is . . . . . . . .
A. 8
B. 4
C. 12
D. 5
Answer: Option A
Solution (By Examveda Team)
a + b = 5c$$\frac{{{a^2} + {b^2} + {c^2} - 1024}}{{ab - bc - ca}} = - 2$$
a2 + b2 + c2 + 2ab - 2bc - 2ac = 1024
(a + b)2 + c(c - 2b - 2a) = 1024
(5c)2 + c[c - 2(b + a)] = 1024
25c2 + c[c - 2 × 5c) = 1024
25c2 - 9c2 = 1024
16c2 = 1024
c2 = 64
c = 8
This Question Belongs to Arithmetic Ability >> Algebra
Join The Discussion