In each of the following questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.
Give answer If
a: x > y
b: x ≤ y
c: x ≥ y
d: x < y
e: Relationship between x and y cannot be established
I.$$x^{2}-x-12= 0$$
II. $$y^{2}+4y+4=0$$
I.$$x^{2} - x - 12 = 0$$
=> $$x^2 - 4x + 3x - 12 = 0$$
=> $$x (x - 4) + 3 (x - 4) = 0$$
=> $$(x - 4) (x + 3) = 0$$
=> $$x = 4 , -3$$
II.$$y^{2} + 4y + 4 = 0$$
=> $$y^2 + 2y + 2y + 4 = 0$$
=> $$y (y + 2) + 2 (y + 2) = 0$$
=> $$(y + 2) (y + 2) = 0$$
=> $$y = -2 , -2$$
Because $$4 > -2$$ and $$-2 > -3$$
$$\therefore$$ No relation can be established.
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