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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.$$12x^{2} + 7x + 1 = 0$$
=> $$12x^2 + 4x + 3x + 1 = 0$$
=> $$4x (3x + 1) + 1 (3x + 1) = 0$$
=> $$(3x + 1) (4x + 1) = 0$$
=> $$x = \frac{-1}{3} , \frac{-1}{4}$$
II.$$6y^{2} + 5y + 1 = 0$$
=> $$6y^2 + 3y + 2y + 1 = 0$$
=> $$3y (2y + 1) + 1 (2y + 1) = 0$$
=> $$(2y + 1) (3y + 1) = 0$$
=> $$y = \frac{-1}{2} , \frac{-1}{3}$$
$$\therefore x \geq y$$
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