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Instructions
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.
Solution
I. $$2x^{2} + 7x + 5 = 0$$
=> $$2x^2 + 2x + 5x + 5 = 0$$
=> $$2x (x + 1) + 5 (x + 1) = 0$$
=> $$(2x + 5) (x + 1) = 0$$
=> $$x = -1 , \frac{-5}{2}$$
II. $$3y^{2} + 5y + 2 = 0$$
=> $$3y^2 + 3y + 2y + 2 = 0$$
=> $$3y (y + 1) + 2 (y + 1) = 0$$
=> $$(3y + 2) (y + 1) = 0$$
=> $$y = -1 , \frac{-2}{3}$$
$$\therefore x \leq y$$
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