Given below are two equations in each question, which you have to solve and give answer
I. ` 3 x^(2) + 7 x + 4 = 0`
II. ` y^(2) + 9 y + 20 = 0`

Options

A. A)If ` x gt y`
B. B)If ` x ge y `
C. C)If ` y gt x`
D. D)If ` y ge x `

Answer

To solve the given equations step by step, we will first tackle each equation separately.

### Step 1: Solve the first equation \(3x^2 + 7x + 4 = 0\)

1. **Identify the coefficients**:
- \(a = 3\), \(b = 7\), \(c = 4\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 7^2 - 4 \cdot 3 \cdot 4 = 49 - 48 = 1 \]

3. **Since the discriminant is positive, we will have two real and distinct roots**.

4. **Use the quadratic formula to find the roots**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-7 \pm \sqrt{1}}{2 \cdot 3} = \frac{-7 \pm 1}{6} \]

5. **Calculate the two roots**:
- First root:
\[ x_1 = \frac{-7 + 1}{6} = \frac{-6}{6} = -1 \]
- Second root:
\[ x_2 = \frac{-7 - 1}{6} = \frac{-8}{6} = -\frac{4}{3} \]

### Step 2: Solve the second equation \(y^2 + 9y + 20 = 0\)

1. **Identify the coefficients**:
- \(a = 1\), \(b = 9\), \(c = 20\)

2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot 20 = 81 - 80 = 1 \]

3. **Since the discriminant is positive, we will have two real and distinct roots**.

4. **Use the quadratic formula to find the roots**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-9 \pm \sqrt{1}}{2 \cdot 1} = \frac{-9 \pm 1}{2} \]

5. **Calculate the two roots**:
- First root:
\[ y_1 = \frac{-9 + 1}{2} = \frac{-8}{2} = -4 \]
- Second root:
\[ y_2 = \frac{-9 - 1}{2} = \frac{-10}{2} = -5 \]

### Summary of Solutions

- For the first equation \(3x^2 + 7x + 4 = 0\), the roots are:
- \(x_1 = -1\)
- \(x_2 = -\frac{4}{3}\)

- For the second equation \(y^2 + 9y + 20 = 0\), the roots are:
- \(y_1 = -4\)
- \(y_2 = -5\)

### Final Comparison

Now, we can compare the roots:
- \(x_1 = -1\) and \(x_2 = -\frac{4}{3}\)
- \(y_1 = -4\) and \(y_2 = -5\)

From the roots:
- \(x_1 = -1 > y_1 = -4\)
- \(x_1 = -1 > y_2 = -5\)
- \(x_2 = -\frac{4}{3} > y_1 = -4\)
- \(x_2 = -\frac{4}{3} > y_2 = -5\)

Thus, we conclude that \(x > y\).