(i) 2x² – 12x + 16 = 0
(ii) 2y² – 20y + 48 = 0
(i) 2x² – 12x + 16 = 0 => 2x² – 8x – 4x + 16 = 0 => 2x (x – 4) – 4 (x – 4) = 0 => (x – 4) (2x – 4) = 0 => x = 4, 2 (ii) 2y² – 20y + 48 = 0 => 2y² – 12 y – 8y + 48 = 0 => 2y(y – 6) – 8(y – 6) = 0 => (y – 6) (2y – 8) = 0 => y = 4, 6 Hence, y ≥ x.
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