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