In the following questions two equations I. and II. are given. You have to solve both the equations and mark the appropriate options
I. ` 12 x ^(2) - 16 x + 5 = 0 `
II. ` 30 y^(2) - 61 y + 30 = 0 `

Options

A. ` x lt y `
B. ` x gt y `
C. ` x le y `
D. ` x le y `

Answer

To solve the equations provided, we will use the quadratic formula for both equations. The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 1: Solve the first equation \(12x^2 - 16x + 5 = 0\)

1. Identify coefficients:
- \(a = 12\)
- \(b = -16\)
- \(c = 5\)

2. Substitute into the quadratic formula:
\[ x = \frac{-(-16) \pm \sqrt{(-16)^2 - 4 \cdot 12 \cdot 5}}{2 \cdot 12} \]

3. Calculate the discriminant:
\[ (-16)^2 = 256 \]
\[ 4 \cdot 12 \cdot 5 = 240 \]
\[ b^2 - 4ac = 256 - 240 = 16 \]

4. Substitute back into the formula:
\[ x = \frac{16 \pm \sqrt{16}}{24} \]
\[ x = \frac{16 \pm 4}{24} \]

5. Calculate the two possible values for \(x\):
- First value:
\[ x_1 = \frac{16 + 4}{24} = \frac{20}{24} = \frac{5}{6} \approx 0.8333 \]
- Second value:
\[ x_2 = \frac{16 - 4}{24} = \frac{12}{24} = \frac{1}{2} = 0.5 \]

### Step 2: Solve the second equation \(30y^2 - 61y + 30 = 0\)

1. Identify coefficients:
- \(a = 30\)
- \(b = -61\)
- \(c = 30\)

2. Substitute into the quadratic formula:
\[ y = \frac{-(-61) \pm \sqrt{(-61)^2 - 4 \cdot 30 \cdot 30}}{2 \cdot 30} \]

3. Calculate the discriminant:
\[ (-61)^2 = 3721 \]
\[ 4 \cdot 30 \cdot 30 = 3600 \]
\[ b^2 - 4ac = 3721 - 3600 = 121 \]

4. Substitute back into the formula:
\[ y = \frac{61 \pm \sqrt{121}}{60} \]
\[ y = \frac{61 \pm 11}{60} \]

5. Calculate the two possible values for \(y\):
- First value:
\[ y_1 = \frac{61 + 11}{60} = \frac{72}{60} = 1.2 \]
- Second value:
\[ y_2 = \frac{61 - 11}{60} = \frac{50}{60} = \frac{5}{6} \approx 0.8333 \]

### Summary of Results:
- Values of \(x\): \(0.8333\) and \(0.5\)
- Values of \(y\): \(1.2\) and \(0.8333\)

### Step 3: Determine the relationship between \(x\) and \(y\)

From the calculated values:
- \(x_1 = 0.8333\) and \(y_2 = 0.8333\)
- \(x_2 = 0.5\) and \(y_1 = 1.2\)

Comparing the values:
- \(y_1 (1.2) > x_1 (0.8333)\)
- \(y_2 (0.8333) = x_1 (0.8333)\)
- \(y_1 (1.2) > x_2 (0.5)\)

Thus, we can conclude that \(y\) is greater than or equal to \(x\).