If $4x^2 + y^2 = 40$ and $xy = 6$, then find the value of $2x + y$.

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Accepted Answer

**Solution:**

1. Use the identity for $(2x + y)^2$:
   $$
   (2x + y)^2 = (2x)^2 + y^2 + 2 \times 2x \times y = 4x^2 + y^2 + 4xy
   $$

2. Substitute the given values:
   $$
   4x^2 + y^2 = 40, \quad xy = 6
   $$
   So,
   $$
   (2x + y)^2 = 40 + 4 \times 6 = 40 + 24 = 64
   $$

3. Take the square root of both sides:
   $$
   2x + y = \pm \sqrt{64} = \pm 8
   $$

4. Since the options are positive, the value is:
   $$
   2x + y = 8
   $$

**Answer:** $8$

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