Instructions

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

Question 126

I. $$30 x^{2} + 11x + 1 = 0$$
II. $$42 y^{2} + 13y + 1 = 0$$

Solution

Statement I : $$30 x^{2} + 11x + 1 = 0$$

=> $$30x^2 + 6x + 5x + 1 = 0$$

=> $$6x (5x + 1) + 1 (5x + 1) = 0$$

=> $$(6x + 1) (5x + 1) = 0$$

=> $$x = \frac{-1}{6} , \frac{-1}{5}$$

Statement II : $$42 y^{2} + 13y + 1 = 0$$

=> $$42y^2 + 7y + 6y + 1 = 0$$

=> $$7y (6y + 1) + 1 (6y + 1) = 0$$

=> $$(7y + 1) (6y + 1) = 0$$

=> $$y = \frac{-1}{7} , \frac{-1}{6}$$

$$\therefore$$ $$x \leq y$$

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I. $$30 x^{2} + 11x + 1 = 0$$II. $$42 y^{2} + 13y + 1 = 0$$