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.
I. $$x^{2}-22x+120=0$$
II. $$y^{2} -26y+168=0$$
I.$$x^{2} - 22x + 120 = 0$$
=> $$x^2 - 10x - 12x + 120 = 0$$
=> $$x (x - 10) - 12 (x - 10) = 0$$
=> $$(x - 10) (x - 12) = 0$$
=> $$x = 10 , 12$$
II.$$y^{2} - 26y + 168 = 0$$
=> $$y^2 - 12y - 14y + 168 = 0$$
=> $$y (y - 12) - 14 (y - 12) = 0$$
=> $$(y - 12) (y - 14) = 0$$
=> $$y = 12 , 14$$
$$\therefore x \leq y$$
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