If `x` and `y` be two positive real numbers such that `4x^(2)+y^(2)=40` and `xy=6` , then find the value od `2x+y` .

Answer

To solve the problem step by step, we start with the given equations:

1. **Given equations:**
- \( 4x^2 + y^2 = 40 \) (Equation 1)
- \( xy = 6 \) (Equation 2)

2. **We need to find the value of \( 2x + y \).**
- Let's denote \( z = 2x + y \).

3. **Using the identity for the square of a binomial:**
- We can express \( z^2 \) as follows:
\[ z^2 = (2x + y)^2 = (2x)^2 + y^2 + 2(2x)(y) = 4x^2 + y^2 + 4xy \]

4. **Substituting the known values into the equation:**
- From Equation 1, we know \( 4x^2 + y^2 = 40 \).
- From Equation 2, we know \( xy = 6 \), so \( 4xy = 4 \cdot 6 = 24 \).

5. **Now substitute these values into the equation for \( z^2 \):**
\[ z^2 = 4x^2 + y^2 + 4xy = 40 + 24 = 64 \]

6. **Taking the square root to find \( z \):**
\[ z = \sqrt{64} = 8 \]

7. **Conclusion:**
- Therefore, the value of \( 2x + y \) is \( 8 \).

### Summary of Steps:
1. Write down the given equations.
2. Define \( z = 2x + y \).
3. Use the binomial square identity to express \( z^2 \).
4. Substitute the known values from the equations into the expression for \( z^2 \).
5. Solve for \( z \) by taking the square root.
6. Conclude with the value of \( 2x + y \).