Instructions

In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and Give answer :

Question 54

I. $$3x^{2} + 5x + 2 = 0$$
II. $$3y^{2} + 18y + 24 = 0$$

Solution

I : 3$$x^{2}$$ + 5x + 2 = 0

=> 3$$x^{2}$$ + 3x + 2x + 2 = 0

=> 3x ( x + 1) + 2 ( x + 1) = 0

=> (3x + 2) (x + 1) = 0

=> x = -1 , -2/3

II : 3$$y^{2}$$ + 18y + 24 = 0

=> 3$$y^{2}$$ + 6y + 12y + 24 = 0

=> 3y (y+2) + 12 (y + 2) = 0

=> (3y + 12) (y + 2) = 0

=> y = -4 , -2

Since, -1 > -4 , -2 && -2/3 > -4 , -2

=> x > y

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I. $$3x^{2} + 5x + 2 = 0$$II. $$3y^{2} + 18y + 24 = 0$$