Question

If 4x^2 + 9y^2 + 16z^2 - 6xy- 12 yz - 8zx = 0 , prove that 2x = 3y = 4z​

Answer 1

$use \: \: \boxed{if \: \: {a}^{2} + {b}^{2} + {c}^{2} = 0 \: \: then, \: \: a = b = c = 0}$

Step-by-step explanation:

$Given, \: \\ 4 {x}^{2} + 9 {y}^{2} + 16 {z}^{2} - 6xy - 12yz - 8zx = 0 \\ On \: \: multiplying \: \: by \: \: 2 \: \: on \: \: both \: \: sides \\ 8 {x}^{2} + 18 {y}^{2} + 32 {z}^{2} - 12xy - 24yz - 16zx = 0 \\ (4 {x}^{2} + 9 {y}^{2} - 12xy ) + (9 {y}^{2} + 16 {z}^{2} - 24yz ) + (16 {z}^{2} + 4 {x}^{2} - 16zx ) = 0 \\ (2x - 3y)^{2} + (3y - 4z)^{2} + (4z - 3x)^{2} = 0 \\ we \: \: know \: \: \boxed{if \: \: {a}^{2} + {b}^{2} + {c}^{2} = 0 \: \: then, \: \: a = b = c = 0} \\ now, \: \: \: 2x - 3y = 3y - 4z = 4z - 3x = 0 \\ so, \: \: \: 2x = 3y = 4z$