Instructions

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

Question 77

I. $$3x^{2}+16x+21=0$$
II. $$6y^{2}+17y+12=0$$

Solution

$$3x^2+16x+21 = 0$$
$$(3x+7)(x+3) = 0$$
$$x = -3, -\frac{7}{3}$$

$$6y^2+17y+12 = 0$$
$$(3y+4)(2y+3) = 0$$
$$y = -\frac{4}{3}, -\frac{3}{2}$$

x < y

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I. $$3x^{2}+16x+21=0$$II. $$6y^{2}+17y+12=0$$

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I. $$3x^{2}+16x+21=0$$
II. $$6y^{2}+17y+12=0$$