1. **Problem statement:** Given the polynomial $x^2 + 4x + 2a$, its zeroes are $a$ and $\frac{2}{a}$. We need to find the value of $a$. 2. **Formula used:** For a quadratic polynomial $x^2 + bx + c$, the sum of zeroes $\alpha + \beta = -b$ and the product of zeroes $\alpha \beta = c$. 3. **Apply the sum of zeroes:** Here, sum of zeroes is $a + \frac{2}{a}$ and it equals $-4$ (since $b=4$). $$a + \frac{2}{a} = -4$$ 4. **Multiply both sides by $a$ to clear the fraction:** $$a^2 + 2 = -4a$$ 5. **Rearrange to form a quadratic equation:** $$a^2 + 4a + 2 = 0$$ 6. **Use the quadratic formula to solve for $a$:** $$a = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times 2}}{2 \times 1} = \frac{-4 \pm \sqrt{16 - 8}}{2} = \frac{-4 \pm \sqrt{8}}{2}$$ 7. **Simplify the square root:** $$\sqrt{8} = 2\sqrt{2}$$ So, $$a = \frac{-4 \pm 2\sqrt{2}}{2} = -2 \pm \sqrt{2}$$ 8. **Check which value satisfies the product of zeroes:** Product of zeroes $= a \times \frac{2}{a} = 2a$ (since $a \neq 0$). From the polynomial, product of zeroes $= 2a$ must equal $2a$ (constant term), so this is consistent. 9. **Conclusion:** The values of $a$ are $-2 + \sqrt{2}$ or $-2 - \sqrt{2}$, which are not among the options given (1, 2, 3, 4). Therefore, none of the options exactly match the value of $a$. However, if the problem expects integer values, none satisfy the condition exactly. **Final answer:** $a = -2 \pm \sqrt{2}$