Given the two equations:

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

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

Determine the correct relationship between $x$ and $y$ based on the roots of these equations.

Choose the correct option:

Question

Given the two equations:

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

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

Determine the correct relationship between $x$ and $y$ based on the roots of these equations.

Choose the correct option:

Accepted Answer

**Solution:**

1. Solve equation I: $3x^2 + 5x + 2 = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a=3$, $b=5$, $c=2$:

$$
   x = \frac{-5 \pm \sqrt{25 - 24}}{6} = \frac{-5 \pm 1}{6}
   $$

So, the roots are:

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

- $x_2 = \frac{-5 - 1}{6} = \frac{-6}{6} = -1$

2. Solve equation II: $3y^2 + 18y + 24 = 0$

Here, $a=3$, $b=18$, $c=24$.

Using the quadratic formula:

$$
   y = \frac{-18 \pm \sqrt{324 - 288}}{6} = \frac{-18 \pm \sqrt{36}}{6} = \frac{-18 \pm 6}{6}
   $$

Roots are:

- $y_1 = \frac{-18 + 6}{6} = \frac{-12}{6} = -2$

- $y_2 = \frac{-18 - 6}{6} = \frac{-24}{6} = -4$

3. Compare roots:

- $x$ roots: $-1, -\frac{2}{3}$

- $y$ roots: $-4, -2$

Since $-\frac{2}{3} > -1 > -2 > -4$, the smallest $x$ root is greater than the largest $y$ root.

Therefore, all $x$ roots are greater than all $y$ roots.

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

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