(2x-y)^2 + (3y-2z)^2 = 0, Find x:y:z.
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(2 x - y)² + (3 y - 2 z)² = 0
(2 x - y)² = - (3 y - 2 z)²
we know that for real numbers, a square of a number is never negative.
Hence 2x - y = 0 => 2x = y => x = y/2
3 y - 2 z = 0 => z = 3/2 * y
ratio : y/2 : y : 3y/2 = 1 : 2 : 3
(2 x - y)² = - (3 y - 2 z)²
we know that for real numbers, a square of a number is never negative.
Hence 2x - y = 0 => 2x = y => x = y/2
3 y - 2 z = 0 => z = 3/2 * y
ratio : y/2 : y : 3y/2 = 1 : 2 : 3
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