Question

If a,b,c,d are four real numbers such that a^2+2b^2+13c^2+5d^2-2ab+4bc-12cd-6d+9=0 then 2a+b+11c+2d is equal to

Answer 1

a, b , c , d are four real numbers in such a way that a² + 2b² + 13c² + 5d² - 2ab + 4bc - 12cd - 6d + 9 = 0

Let's start to arrange it first ,
a² + 2b² + 13c² + 5d² - 2ab + 4bc - 12cd - 6d + 9 = 0
⇒( a² + b² - 2ab ) + (b² + 4c² + 4bc ) + (9c² + 4d² -12cd) + (d² - 6d + 9) = 0
⇒(a - b)² + (b + 2c)² + (3c - 2d)² + (d - 3)² = 0
LHS will be RHS only when (a - b) = 0 ⇒a = b
(b + 2c ) = 0 ⇒b = -2c
(3c - 2d) = 0 ⇒3c = 2d
(d - 3) = 0 ⇒ d = 3 , put it in 3c = 2d then, c = 2
put c = 2 in b = -2c = -4
put b = -4c in a = b = -4

Now, 2a + b + 11c + 2d = 2 × -4 + (-4) + 11 × 2 + 2 × 3
= -8 - 4 + 22 + 6
= -12 + 22 + 6
= 16