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Instructions
In each of these questions two equations numbered I and II are given. You have to solve both the equations and –
Give answer a: if x < y
Give answer b: if x ≤ y
Give answer c: if x > y
Give answer d: if x ≥ y
Give answer e: if x = y or the relationship cannot be established.
Solution
I.$$x^{2} + 7x + 12 = 0$$
=> $$x^2 + 3x + 4x + 12 = 0$$
=> $$x (x + 3) + 4 (x + 3) = 0$$
=> $$(x + 3) (x + 4) = 0$$
=> $$x = -3 , -4$$
II.$$y^{2} + 6y + 8 = 0$$
=> $$y^2 + 4y + 2y + 8 = 0$$
=> $$y (y + 4) + 2 (y + 4) = 0$$
=> $$(y + 4) (y + 2) = 0$$
=> $$y = -4 , -2$$
Because $$-2 > -4$$ and $$-3 > -4$$
Therefore, no relation can be established.
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