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 128

I.$$x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$$
II.$$y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$$

Solution

Statement I : $$x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$$

=> $$x (x - 2) - \sqrt{5} (x - 2) = 0$$

=> $$(x - \sqrt{5}) (x - 2) = 0$$

=> $$x = \sqrt{5} , 2$$

Statement II : $$y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$$

=> $$y (y - \sqrt{3}) - \sqrt{2} (y - \sqrt{3}) = 0$$

=> $$(y - \sqrt{2}) (y - \sqrt{3}) = 0$$

=> $$y = \sqrt{2} , \sqrt{3}$$

$$\therefore$$ $$x > y$$

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I.$$x^{2}-2x-\sqrt{5}x+2\sqrt{5}=0$$II.$$y^{2}-\sqrt{3}y-\sqrt{2}y+\sqrt{6}=0$$