1. The problem is to find the value of $k$ for which the quadratic equation $$x^2 + (k+8)x + 9k = 0$$ has a real double root. 2. A quadratic equation $$ax^2 + bx + c = 0$$ has a real double root if its discriminant $$\Delta = b^2 - 4ac$$ is zero. 3. For the given equation, identify coefficients: $$a = 1$$, $$b = k+8$$, and $$c = 9k$$. 4. Write the discriminant condition for a double root: $$\Delta = (k+8)^2 - 4 \times 1 \times 9k = 0$$ 5. Expand and simplify: $$ (k+8)^2 - 36k = 0 $$ $$ k^2 + 16k + 64 - 36k = 0 $$ $$ k^2 - 20k + 64 = 0 $$ 6. Solve the quadratic equation for $k$: $$ k = \frac{20 \pm \sqrt{(-20)^2 - 4 \times 1 \times 64}}{2} = \frac{20 \pm \sqrt{400 - 256}}{2} = \frac{20 \pm \sqrt{144}}{2} $$ 7. Calculate the roots: $$ k = \frac{20 \pm 12}{2} $$ 8. So, $$ k_1 = \frac{20 + 12}{2} = 16 $$ $$ k_2 = \frac{20 - 12}{2} = 4 $$ 9. Therefore, the quadratic equation has a real double root when $$k = 4$$ or $$k = 16$$.