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II. 3y2 + 20y + 33 = 01). If x > y2). If x < y3). If x ≥ y4). If x ≤ y |
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Answer» ⇒ 2x2 + 4x + 5X + 10 = 0 ⇒ 2x(x + 2) + 5(x + 2) = 0 ⇒ (2x + 5)(x + 2) = 0 x = -5/2, -2 3y2 + 20y + 33 = 0 ⇒ 3y2 + 9Y + 11y + 33 = 0 ⇒ 3y(y + 3) + 11(y + 3) = 0 ⇒ (3y + 11)(y + 3) = 0 y = -11/3, -3 Thus, x = -5/2, -2 and y = -11/3, -3 Comparing these values of x and y, we get x > y |
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