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