Instructions

In these questions, two equations numbered I and II are given. You have to solve both the equations and select the appropriate option.

Question 49

I. $$2x^{2}+19x+45=0$$
II. $$2y^{2}+11y+12=0$$

Solution

$$2x^2+19x+45 = 0$$
$$(2x+9)(x+5) = 0$$
$$x = -5, -\frac{9}{2}$$

$$2y^2+11y+12 = 0$$
$$(2y+3)(y+4) = 0$$
$$y = -4, -\frac{3}{2}$$

x < y

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I. $$2x^{2}+19x+45=0$$II. $$2y^{2}+11y+12=0$$