Instructions

In each of the following questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.
Give answer If
a: x > y
b: x ≤ y
c: x ≥ y
d: x < y
e: Relationship between x and y cannot be established

Question 64

I. $$x^{2}+11x+28=0$$
II. $$5y^{2}+27y+28=0$$

Solution

I.$$x^{2} + 11x + 28 = 0$$

=> $$x^2 + 7x + 4x + 28 = 0$$

=> $$x (x + 7) + 4 (x + 7) = 0$$

=> $$(x + 7) (x + 4) = 0$$

=> $$x = -7 , -4$$

II.$$5y^{2} + 27y + 28 = 0$$

=> $$5y^2 + 20y + 7y + 28 = 0$$

=> $$5y (y + 4) + 7 (y + 4) = 0$$

=> $$(y + 4) (5y + 7) = 0$$

=> $$y = -4 , \frac{-7}{5}$$

$$\therefore x \leq y$$

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I. $$x^{2}+11x+28=0$$II. $$5y^{2}+27y+28=0$$

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I. $$x^{2}+11x+28=0$$
II. $$5y^{2}+27y+28=0$$