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 127

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

Solution

I. $$x^{2}- x - \sqrt{2}x + \sqrt{2}=0$$

=> $$x (x - 1) - \sqrt{2} (x - 1) = 0$$

=> $$(x - \sqrt{2}) (x - 1) = 0$$

=> $$x = \sqrt{2} , 1$$

II. $$y^{2}-3y+2=0$$

=> $$y^2 - 2y - y + 2 = 0$$

=> $$y (y - 2) - 1 (y - 2) = 0$$

=> $$(y - 2) (y - 1) = 0$$

=> $$y = 1 , 2$$

$$\therefore$$ No relation established.

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