1. **State the problem:** We need to show that for all values of $k$, the quadratic equation $$f(x) = 2x^2 + (k+8)x + k = 0$$ has distinct real roots. 2. **Recall the formula for roots of a quadratic:** A quadratic equation $ax^2 + bx + c = 0$ has distinct real roots if its discriminant $$\Delta = b^2 - 4ac > 0.$$ 3. **Identify coefficients:** Here, $a = 2$, $b = k+8$, and $c = k$. 4. **Write the discriminant:** $$\Delta = (k+8)^2 - 4 \times 2 \times k = (k+8)^2 - 8k.$$ 5. **Expand and simplify:** $$ (k+8)^2 - 8k = k^2 + 16k + 64 - 8k = k^2 + 8k + 64.$$ 6. **Analyze the discriminant:** We want to show $$k^2 + 8k + 64 > 0$$ for all real $k$. 7. **Complete the square:** $$k^2 + 8k + 64 = (k^2 + 8k + 16) + 48 = (k+4)^2 + 48.$$ 8. **Interpretation:** Since $(k+4)^2 \geq 0$ for all $k$ and $48 > 0$, the sum is always positive. 9. **Conclusion:** Therefore, $$\Delta > 0$$ for all real $k$, so the quadratic equation has distinct real roots for all values of $k$. **Final answer:** The equation $f(x) = 0$ has distinct real roots for every real value of $k$ because its discriminant is always positive.