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. $$2x^{2}+17x+36=0$$
II. $$3y^{2} +20y+33=0$$
I.$$2x^{2} + 17x + 36 = 0$$
=> $$2x^2 + 8x + 9x + 36 = 0$$
=> $$2x (x + 4) + 9 (x + 4) = 0$$
=> $$(x + 4) (2x + 9) = 0$$
=> $$x = -4 , \frac{-9}{2}$$
II.$$3y^{2} + 20y + 33 = 0$$
=> $$3y^2 + 9y + 11y + 33 = 0$$
=> $$3y (y + 3) + 11 (y + 3) = 0$$
=> $$(y + 3) (3y + 11) = 0$$
=> $$y = -3 , \frac{-11}{3}$$
$$\therefore x < y$$
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