Given the two quadratic equations:

(I) $12x^2 + 46x + 42 = 0$

(II) $3y^2 - 16y + 21 = 0$

If $x$ and $y$ are the roots of equations (I) and (II) respectively, determine the relationship between $x$ and $y$.

Question

Given the two quadratic equations:

(I) $12x^2 + 46x + 42 = 0$

(II) $3y^2 - 16y + 21 = 0$

If $x$ and $y$ are the roots of equations (I) and (II) respectively, determine the relationship between $x$ and $y$.

Accepted Answer

**Solution:**

1. First, solve equation (I):
   $$12x^2 + 46x + 42 = 0$$

Calculate the discriminant ($D$):
   $$D = b^2 - 4ac = 46^2 - 4 	imes 12 	imes 42 = 2116 - 2016 = 100$$

Since $D > 0$, roots are real and distinct.

Roots of (I) are:
   $$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-46 \pm 10}{24}$$

So,
   $$x_1 = \frac{-46 + 10}{24} = \frac{-36}{24} = -1.5$$
   $$x_2 = \frac{-46 - 10}{24} = \frac{-56}{24} = -2.333...$$

2. Now, solve equation (II):
   $$3y^2 - 16y + 21 = 0$$

Calculate the discriminant ($D$):
   $$D = (-16)^2 - 4 	imes 3 	imes 21 = 256 - 252 = 4$$

Roots are real and distinct.

Roots of (II) are:
   $$y = \frac{16 \pm 2}{6}$$

So,
   $$y_1 = \frac{16 + 2}{6} = \frac{18}{6} = 3$$
   $$y_2 = \frac{16 - 2}{6} = \frac{14}{6} = 2.333...$$

3. Compare the roots:
   - Roots of $x$ are approximately $-1.5$ and $-2.333$.
   - Roots of $y$ are approximately $3$ and $2.333$.

Since both roots of $x$ are less than both roots of $y$, we conclude:

$$x < y$$

**Answer:** (c) $x < y$

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