Question

If a^2 + b^2 + 4c^2 = 2(a + b - 2c) - 3 and a, b, c are real such that (a^99 + b^999 + c^2) = k + 1/4 then k^2 =

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Answer 1

Heya User,

Nice Question... Ohk Let's see :->

--> a² + b² + (-2c)² = 2a + 2b - 4c - 3
=> [ a² - 2a + 1 ] + [ b² - 2b + 1 ] + [ 4c² + 4c + 1 ] = 0
=> ( a - 1 )² + ( b - 1 )² + ( 2c + 1 )² = 0 ;

Now, considering that a, b, c are real no.s and that any k∈R,
--> k² ≥ 0

--> which gives :-> a - 1 = 0 ; b - 1 = 0 ; 2c + 1 = 0

And hence, from there follows :-> k = 2 and hence, k² = 4

Another approach would be considering the inequalities but anyways, this'd do just fine... xD

--> Hope you liked the soln.