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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 cannot be established.
Solution
I.$$3x^{2} + 10x + 8 = 0$$
=> $$3x^2 + 6x + 4x + 8 = 0$$
=> $$3x (x + 2) + 4 (x + 2) = 0$$
=> $$(x + 2) (3x + 4) = 0$$
=> $$x = -2 , \frac{-4}{3}$$
II.$$3y^{2} + 7y + 4 = 0$$
=> $$3y^2 + 3y + 4y + 4 = 0$$
=> $$3y (y + 1) + 4 (y + 1) = 0$$
=> $$(y + 1) (3y + 4) = 0$$
=> $$y = -1 , \frac{-4}{3}$$
$$\therefore x \leq y$$
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