In the following 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 between x and y cannot be established.
I. $$x^{2}+6x+9=0$$
II. $$y^{2}-y-20=0$$
I. $$x^{2} + 6x + 9 = 0$$
=> $$x^2 + 3x + 3x + 9 = 0$$
=> $$x (x + 3) + 3 (x + 3) = 0$$
=> $$(x + 3) (x + 3) = 0$$
=> $$x = -3 , -3$$
II. $$y^{2} - y - 20 = 0$$
=> $$y^2 + 4y - 5y - 20 = 0$$
=> $$y (y + 4) - 5 (y + 4) = 0$$
=> $$(y + 4) (y - 5) = 0$$
=> $$y = -4 , 5$$
Because $$-3 > -4$$ and $$5 > -3$$
No relation can be established.
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