Given the two quadratic equations:

Quantity I: $$\frac{x^2}{5} + x + \frac{4}{5} = 0$$

Quantity II: $$3y^2 + 4y + 1 = 0$$

Let *x* and *y* be the roots of the respective equations. Compare Quantity I and Quantity II where:

- Quantity I = *x*
- Quantity II = *y*

Determine the correct relationship between Quantity I and Quantity II.

Question

Given the two quadratic equations:

Quantity I: $$\frac{x^2}{5} + x + \frac{4}{5} = 0$$

Quantity II: $$3y^2 + 4y + 1 = 0$$

Let *x* and *y* be the roots of the respective equations. Compare Quantity I and Quantity II where:

- Quantity I = *x*
- Quantity II = *y*

Determine the correct relationship between Quantity I and Quantity II.

Accepted Answer

**Solution:**

1. First, simplify the equation for Quantity I:

$$\frac{x^2}{5} + x + \frac{4}{5} = 0$$

Multiply through by 5 to clear the denominator:

$$x^2 + 5x + 4 = 0$$

2. Factorize the quadratic:

$$x^2 + 4x + x + 4 = 0$$

$$x(x + 4) + 1(x + 4) = 0$$

$$(x + 4)(x + 1) = 0$$

So, the roots are:

$$x = -1, -4$$

3. Now, consider the equation for Quantity II:

$$3y^2 + 4y + 1 = 0$$

4. Factorize this quadratic:

$$3y^2 + 3y + y + 1 = 0$$

$$3y(y + 1) + 1(y + 1) = 0$$

$$(y + 1)(3y + 1) = 0$$

Hence, the roots are:

$$y = -1, -\frac{1}{3}$$

5. Now, compare the roots:

- Quantity I roots: $-4, -1$

- Quantity II roots: $-1, -\frac{1}{3}$

6. The largest root of Quantity I is $-1$, and the smallest root of Quantity II is $-1$. The other roots are $-4$ for Quantity I and $-\frac{1}{3}$ for Quantity II.

7. Since $-4 < -1 < -\frac{1}{3}$, the roots of Quantity I are less than or equal to the roots of Quantity II.

**Answer:** Quantity I $\leq$ Quantity II

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