`2x^2+6xy+5y^2=1, then`

Options

A. `|x|le sqrt(5)`
B. `|x|ge sqrt(5)`
C. `y^(2) le 2`
D. `y^(2) le 4`

Explanation

To solve the equation \(2x^2 + 6xy + 5y^2 = 1\) and determine the conditions for \(x\) and \(y\), we will follow these steps:

### Step 1: Rewrite the equation as a quadratic in \(x\)
We can rearrange the given equation:
\[ 2x^2 + 6xy + 5y^2 - 1 = 0 \]
This is a quadratic equation in \(x\) of the form \(ax^2 + bx + c = 0\) where:
- \(a = 2\)
- \(b = 6y\)
- \(c = 5y^2 - 1\)

### Step 2: Apply the discriminant condition
For the quadratic equation to have real solutions for \(x\), the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values of \(a\), \(b\), and \(c\):
\[ D = (6y)^2 - 4(2)(5y^2 - 1) \geq 0 \]
Calculating the discriminant:
\[ D = 36y^2 - 8(5y^2 - 1) \geq 0 \]
\[ D = 36y^2 - 40y^2 + 8 \geq 0 \]
\[ D = -4y^2 + 8 \geq 0 \]

### Step 3: Solve the inequality
Rearranging the inequality:
\[ -4y^2 + 8 \geq 0 \]
\[ 4y^2 \leq 8 \]
\[ y^2 \leq 2 \]
Taking the square root:
\[ |y| \leq \sqrt{2} \]

### Step 4: Rewrite the equation as a quadratic in \(y\)
Now, we rewrite the original equation as a quadratic in \(y\):
\[ 5y^2 + 6xy + 2x^2 - 1 = 0 \]
Here, we identify:
- \(a = 5\)
- \(b = 6x\)
- \(c = 2x^2 - 1\)

### Step 5: Apply the discriminant condition for \(y\)
For this quadratic to have real solutions for \(y\), we again require:
\[ D = b^2 - 4ac \geq 0 \]
Substituting the values:
\[ D = (6x)^2 - 4(5)(2x^2 - 1) \geq 0 \]
Calculating the discriminant:
\[ D = 36x^2 - 20(2x^2 - 1) \geq 0 \]
\[ D = 36x^2 - 40x^2 + 20 \geq 0 \]
\[ D = -4x^2 + 20 \geq 0 \]

### Step 6: Solve the inequality for \(x\)
Rearranging:
\[ -4x^2 + 20 \geq 0 \]
\[ 4x^2 \leq 20 \]
\[ x^2 \leq 5 \]
Taking the square root:
\[ |x| \leq \sqrt{5} \]

### Final Result
From our calculations, we find:
- The condition for \(y\) is \(|y| \leq \sqrt{2}\)
- The condition for \(x\) is \(|x| \leq \sqrt{5}\)