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 cannot be established.
I. $$5x^{2}+29x+20=0$$
II. $$25y^{2}+25y+6=0$$
I.$$5x^{2} + 29x + 20 = 0$$
=> $$5x^2 + 25x + 4x + 20 = 0$$
=> $$5x (x + 5) + 4 (x + 5) = 0$$
=> $$(x + 5) (5x + 4) = 0$$
=> $$x = -5 , \frac{-4}{5}$$
II.$$25y^{2} + 25y + 6 = 0$$
=> $$25y^2 + 10y + 15y + 6 = 0$$
=> $$5y (5y + 2) + 3 (5y + 2) = 0$$
=> $$(5y + 3) (5y + 2) = 0$$
=> $$y = \frac{-3}{5} , \frac{-2}{5}$$
Therefore $$x < y$$
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