Question
I. 4x2 – 53x – 105 = 0 II.
3y2 – 25y + 48 = 0Solution
From I: 4x2 – 53x – 105 = 0 Or, 4x2 – 60x + 7x – 105 = 0 Or, 4x × (x – 15) + 7 × (x – 15) = 0 Or, (4x + 7) × (x – 15) = 0 So, x = – (7/4) = – 1.75 or, x = 15 From II: 3y2 – 25y + 48 = 0 Or, 3y2 – 9y – 16y + 48 = 0 Or, 3y × (y – 3) – 16 × (y – 3) = 0 Or, (y – 3) × (3y – 16) = 0 So, y = 3 or y = (16/3) ~ 5.33 
So, the relationship cannot be established. Hence, option c.
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