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 157

I. $${x^2}$$ + 28x + 192 = 0
II. $${y^2}$$ + 16y + 48 = 0

Solution

I.$$x^{2} + 28x + 192 = 0$$

=> $$x^2 + 16x + 12x + 192 = 0$$

=> $$x (x + 16) + 12 (x + 16) = 0$$

=> $$(x + 16) (x + 12) = 0$$

=> $$x = -16 , -12$$

II.$$y^{2} + 16y + 48 = 0$$

=> $$y^2 + 12y + 4y + 48 = 0$$

=> $$y (y + 12) + 4 (y + 12) = 0$$

=> $$(y + 12) (y + 4) = 0$$

=> $$y = -12 , -4$$

$$\therefore x \leq y$$

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I. $${x^2}$$ + 28x + 192 = 0II. $${y^2}$$ + 16y + 48 = 0