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 129

I.$$ x^{2}+12x+36=0$$
II.$$y^{2}=16$$

Solution

Statement 1 : $$x^2 + 12x + 36 = 0$$

=> $$x^2 + 2.x.6 + 6^2 = 0$$

=> $$(x + 6)^2 = 0$$

=> $$x = -6$$

Statement II : $$y^2 = 16$$

=> $$(y)^2 = (\pm 4)^2$$

=> $$y = \pm 4$$

$$\therefore$$ $$x < y$$

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I.$$ x^{2}+12x+36=0$$II.$$y^{2}=16$$