In each of the following questions, two equations are given. You have to solve them and give answer(निम्नलिखित प्रत्येक प्रश्न में दो समीकरण दिए गए हैं। आपको उन्हें हल करना है और उत्तर देना है)
I. `3x^(2) - 16 x+21=0`
II. `6y^2 +25 y +21 =0`

Options

A. `xgty`
B. `xgey`
C. `x lty`
D. `xley`

Answer

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

### Step 1: Solve the first equation \(3x^2 - 16x + 21 = 0\)

1. **Identify the coefficients**: Here, \(a = 3\), \(b = -16\), and \(c = 21\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (-16)^2 - 4 \cdot 3 \cdot 21 = 256 - 252 = 4 \]
3. **Find the roots using the quadratic formula**:
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{16 \pm \sqrt{4}}{2 \cdot 3} = \frac{16 \pm 2}{6} \]
- First root:
\[ x_1 = \frac{16 + 2}{6} = \frac{18}{6} = 3 \]
- Second root:
\[ x_2 = \frac{16 - 2}{6} = \frac{14}{6} = \frac{7}{3} \]

### Step 2: Solve the second equation \(6y^2 + 25y + 21 = 0\)

1. **Identify the coefficients**: Here, \(a = 6\), \(b = 25\), and \(c = 21\).
2. **Calculate the discriminant**:
\[ D = b^2 - 4ac = (25)^2 - 4 \cdot 6 \cdot 21 = 625 - 504 = 121 \]
3. **Find the roots using the quadratic formula**:
\[ y = \frac{-b \pm \sqrt{D}}{2a} = \frac{-25 \pm \sqrt{121}}{2 \cdot 6} = \frac{-25 \pm 11}{12} \]
- First root:
\[ y_1 = \frac{-25 + 11}{12} = \frac{-14}{12} = -\frac{7}{6} \]
- Second root:
\[ y_2 = \frac{-25 - 11}{12} = \frac{-36}{12} = -3 \]

### Step 3: Compare the roots

- The roots of the first equation are \(x_1 = 3\) and \(x_2 = \frac{7}{3}\).
- The roots of the second equation are \(y_1 = -\frac{7}{6}\) and \(y_2 = -3\).

### Step 4: Determine the relationship between \(x\) and \(y\)

- Since both roots of \(y\) are negative and both roots of \(x\) are positive, we can conclude that:
- \(x > y\) for all combinations of roots from both equations.

### Final Answer

Since \(x\) is greater than \(y\), the answer is **1** (indicating \(x > y\)).

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