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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 If
a: x > y
b: x ≥ y
c: x < y
d: x ≤ y
e: x = y or the relationship can not be established
Solution
I : $$x^{2} + 5x + 6 = 0$$
=> $$x^2 + 2x + 3x + 6 = 0$$
=> $$x (x + 2) + 3 (x + 2) = 0$$
=> $$(x + 2) (x + 3) = 0$$
=> $$x = -2 , -3$$
II : $$y^{2} + 3y + 2 = 0$$
=> $$y^2 + 2y + y + 2 = 0$$
=> $$y (y + 2) + 1 (y + 2) = 0$$
=> $$(y + 2) (y + 1) = 0$$
=> $$y = -1 , -2$$
$$\therefore x \leq y$$
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