Instructions

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

Question 53

I. $$3x^{2} + 20x + 25 = 0$$
II. $$3y^{2} + 14y + 8 =0$$

Solution

I : 3$$x^{2}$$ + 20x + 25 = 0

=> 3$$x^{2}$$ + 15x + 5x + 25 = 0

=> 3x ( x + 5) + 5 ( x + 5) = 0

=> (3x + 5) (x + 5) = 0

=> x = -5 , -5/3

II : 3$$y^{2}$$ + 14y + 8 = 0

=> 3$$y^{2}$$ + 12y + 2y + 8 = 0

=> 3y (y+4) + 2 (y + 4) = 0

=> (3y + 2) (y + 4) = 0

=> y = --2/3 , -4

Since, -2/3 > -5/3 > -4

=> Relation cannot be determined.

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I. $$3x^{2} + 20x + 25 = 0$$II. $$3y^{2} + 14y + 8 =0$$