Given below are two equations in each question, which you have to solve and give answer
I. ` 2 x^(2) - 5 x + 2 = 0`
II. ` 2 y^(2) - 9 y + 7 = 0`

Options

A. If ` x gt y`
B. If ` x ge y `
C. If ` y gt x`
D. If x = y or no relation can be established

Answer

To solve the given equations step by step, we will start with the first equation and then move on to the second equation.

### Step 1: Solve the first equation `2x^2 - 5x + 2 = 0`

1. **Identify the coefficients**:
- Here, \(a = 2\), \(b = -5\), and \(c = 2\).

2. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot 2}}{2 \cdot 2} \]

3. **Calculate the discriminant**:
\[ (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9 \]

4. **Substitute back into the formula**:
\[ x = \frac{5 \pm \sqrt{9}}{4} = \frac{5 \pm 3}{4} \]

5. **Find the two possible values for x**:
- First value:
\[ x_1 = \frac{5 + 3}{4} = \frac{8}{4} = 2 \]
- Second value:
\[ x_2 = \frac{5 - 3}{4} = \frac{2}{4} = 0.5 \]

### Step 2: Solve the second equation `2y^2 - 9y + 7 = 0`

1. **Identify the coefficients**:
- Here, \(a = 2\), \(b = -9\), and \(c = 7\).

2. **Use the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Plugging in the values:
\[ y = \frac{-(-9) \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 7}}{2 \cdot 2} \]

3. **Calculate the discriminant**:
\[ (-9)^2 - 4 \cdot 2 \cdot 7 = 81 - 56 = 25 \]

4. **Substitute back into the formula**:
\[ y = \frac{9 \pm \sqrt{25}}{4} = \frac{9 \pm 5}{4} \]

5. **Find the two possible values for y**:
- First value:
\[ y_1 = \frac{9 + 5}{4} = \frac{14}{4} = 3.5 \]
- Second value:
\[ y_2 = \frac{9 - 5}{4} = \frac{4}{4} = 1 \]

### Step 3: Compare the values of x and y

- The values of \(x\) are \(0.5\) and \(2\).
- The values of \(y\) are \(1\) and \(3.5\).

Now, we compare:
- \(0.5 < 1\) (so \(x < y\))
- \(0.5 < 3.5\) (so \(x < y\))
- \(2 > 1\) (so \(x > y\))
- \(2 < 3.5\) (so \(x < y\))

From these comparisons, we can conclude that in all cases where \(x\) is less than \(y\), we can state that \(x < y\) or equivalently \(y > x\).

### Final Answer:
The relation between \(x\) and \(y\) is \(y > x\).

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