Given the two quadratic equations:

(I) $3x^2 + 5x + 2 = 0$

(II) $3y^2 + 18y + 24 = 0$

Determine the correct relationship between $x$ and $y$.

Question

Given the two quadratic equations:

(I) $3x^2 + 5x + 2 = 0$

(II) $3y^2 + 18y + 24 = 0$

Determine the correct relationship between $x$ and $y$.

Accepted Answer

**Solution:**

1. Solve equation (I):

$$3x^2 + 5x + 2 = 0$$

Using the quadratic formula:

$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{25 - 24}}{6} = \frac{-5 \pm 1}{6}$$

So,
- $x_1 = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3}$
- $x_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1$

Thus, $x$ can be either $-\frac{2}{3}$ or $-1$.

2. Solve equation (II):

$$3y^2 + 18y + 24 = 0$$

Divide entire equation by 3:

$$y^2 + 6y + 8 = 0$$

Apply quadratic formula:

$$y = \frac{-6 \pm \sqrt{36 - 32}}{2} = \frac{-6 \pm 2}{2}$$

So,
- $y_1 = \frac{-6 + 2}{2} = \frac{-4}{2} = -2$
- $y_2 = \frac{-6 - 2}{2} = \frac{-8}{2} = -4$

Therefore, $y$ can be either $-2$ or $-4$.

3. Compare possible values of $x$ and $y$:

- $x_1 = -\frac{2}{3} \approx -0.666$, $x_2 = -1$
- $y_1 = -2$, $y_2 = -4$

Since both $y$ values are less than both $x$ values, for all combinations:

- $x_1 > y_1$, $x_1 > y_2$
- $x_2 > y_1$, $x_2 > y_2$

Hence, $x$ is always greater than $y$.

**Answer:** (a) $x > y$

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