In each of the following questions, two equations I and II have been given. Solve these questions and answer
(1)if x < y
(2) if x ≤ y
(3) if x = y or the relation cannot be established
(4) if ≥ y
(5) if x > y
Statement I : $$9x^{2}+3x-2=0$$
=> $$9x^2 + 6x - 3x - 2 = 0$$
=> $$3x (3x + 2) - 1 (3x + 2) = 0$$
=> $$(3x - 1) (3x + 2) = 0$$
=> $$x = \frac{1}{3} , \frac{-2}{3}$$
Statement II : $$8y^{2}+6y+1=0$$
=> $$8y^2 + 4y + 2y + 1 = 0$$
=> $$4y (2y + 1) + 1 (2y + 1) = 0$$
=> $$(4y + 1) (2y + 1) = 0$$
=> $$y = \frac{-1}{4} , \frac{-1}{2}$$
$$\therefore$$ No relation can be established.
Create a FREE account and get: