Instructions

In the following questions two equations numbered I and II are given. You have to solve both equations and
Give answer If

a. x ˃ y
b. x ≥ y
c. x ˂ y
d. x ≤ y
e. x = y or the relationship cannot be established

Question 156

I. $${x^2}$$ - 7x + 10 = 0
II. $${y^2}$$ + 11y + 10 = 0

Solution

I.$$x^{2} - 7x + 10 = 0$$

=> $$x^2 - 5x - 2x + 10 = 0$$

=> $$x (x - 5) - 2 (x - 5) = 0$$

=> $$(x - 5) (x - 2) = 0$$

=> $$x = 5 , 2$$

II.$$y^{2} + 11y + 10 = 0$$

=> $$y^2 + 10y + y + 10 = 0$$

=> $$y (y + 10) + 1 (y + 10) = 0$$

=> $$(y + 10) (y + 1) = 0$$

=> $$y = -10 , -1$$

$$\therefore x > y$$

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I. $${x^2}$$ - 7x + 10 = 0II. $${y^2}$$ + 11y + 10 = 0