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II. 35y2 + 48y - 27 = 01). x > y2). x < y3). x ≥ y4). x ≤ y

Answer»

I. 4x2 + 4√3x + 3 = 0

⇒ 4x2 + 2√3x + 2√3x + 3 = 0

⇒ 2x(2x + √3) + √3(2x + √3) = 0

⇒ (2x + √3)2 = 0

∴ x =-√3/2

II. 35y2 + 48y - 27 = 0

⇒ 35y2 + 63y - 15Y - 27 = 0

⇒ 7y(5Y + 9) - 3(5y + 9) = 0

⇒ (5y + 9)(7y - 3) = 0

∴ y =-9/5 or y = 3/7

If x = -√3/2, x > y for y = -9/5

If x = -√3/2, x < y for y = 3/7

∴ This shows that relationship cannot be ESTABLISHED between x and y



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