Instructions

In the following questions two equations numbered I and II are given. You have to solve both the equations and
Give answer a: if x >y
Give answer b: if x ≥ y
Give answer c: if x < y
Give answer d: if x ≤ y
Give answer e: if x = y or the relationship between x and y cannot be established.

Question 16

I. $$x^{2}-4x-12=0$$
II. $$y^{2}-5y-14=0$$

Solution

I.$$x^{2} - 4x - 12 = 0$$

=> $$x^2 + 2x - 6x - 12 = 0$$

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

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

=> $$x = -2 , 6$$

II.$$y^{2} - 5y - 14 = 0$$

=> $$y^2 + 2y - 7y - 14 = 0$$

=> $$y (y + 2) - 7 (y + 2) = 0$$

=> $$(y + 2) (y - 7) = 0$$

=> $$y = -2 , 7$$

Because $$7 > -2$$ and $$6 > -2$$

No relation can be established.


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