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Instructions
In the following questions two equations numbered I and II are given. You have to solve both the equations and —
Give answer a: if × > y
Give answer b: if × ≥ y
Give answer c: if × < y
Give answer d: if × ≤ y
Give answer e: if × = y or the relationship cannot be established.
Solution
I. $$3x^{2} + 13x+ 12 = 0$$
=> $$3x^2 + 9x + 4x + 12 = 0$$
=> $$3x (x + 3) + 4(x + 3) = 0$$
=> $$(3x + 4) (x + 3) = 0$$
=> $$x = \frac{-4}{3} , -3$$
II. $$2y^{2} + 15y+ 27= 0$$
=> $$2y^2 + 6y + 9y + 27 = 0$$
=> $$(2y + 9) (y + 3) = 0$$
=> $$y = \frac{-9}{2} , -3$$
$$\therefore x \geq y$$
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