1. **State the problem:** We have the quadratic equation $k + 9 = (6 - x)(2x + 1)$ and need to find the range of values of $k$ for which the equation has two distinct real roots. 2. **Rewrite the equation:** Expand the right side: $$ (6 - x)(2x + 1) = 6 \cdot 2x + 6 \cdot 1 - x \cdot 2x - x \cdot 1 = 12x + 6 - 2x^2 - x = -2x^2 + 11x + 6 $$ So the equation becomes: $$ k + 9 = -2x^2 + 11x + 6 $$ 3. **Bring all terms to one side:** $$ -2x^2 + 11x + 6 - k - 9 = 0 $$ $$ -2x^2 + 11x - (k + 3) = 0 $$ Rewrite as: $$ -2x^2 + 11x - (k + 3) = 0 $$ 4. **Identify coefficients:** $$ a = -2, \quad b = 11, \quad c = -(k + 3) $$ 5. **Condition for two distinct real roots:** The discriminant $\Delta$ must be positive: $$ \Delta = b^2 - 4ac > 0 $$ Substitute: $$ 11^2 - 4(-2)(-(k + 3)) > 0 $$ $$ 121 - 8(k + 3) > 0 $$ 6. **Solve inequality:** $$ 121 - 8k - 24 > 0 $$ $$ 97 - 8k > 0 $$ $$ -8k > -97 $$ $$ \cancel{-8}k < \cancel{-97} \quad \Rightarrow \quad k < \frac{97}{8} $$ 7. **Range of $k$ for two distinct real roots:** $$ k < \frac{97}{8} = 12.125 $$ --- **Part (b):** Let $k$ be the largest possible integer satisfying the inequality, so: $$ k = 12 $$ 8. **Solve the quadratic equation for $k=12$:** $$ -2x^2 + 11x - (12 + 3) = 0 $$ $$ -2x^2 + 11x - 15 = 0 $$ 9. **Use quadratic formula:** $$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ $$ x = \frac{-11 \pm \sqrt{11^2 - 4(-2)(-15)}}{2(-2)} $$ $$ x = \frac{-11 \pm \sqrt{121 - 120}}{-4} $$ $$ x = \frac{-11 \pm \sqrt{1}}{-4} $$ 10. **Calculate roots:** $$ x_1 = \frac{-11 + 1}{-4} = \frac{-10}{-4} = \frac{5}{2} = 2.5 $$ $$ x_2 = \frac{-11 - 1}{-4} = \frac{-12}{-4} = 3 $$ **Final answers:** - (a) The range of $k$ for two distinct real roots is: $$ k < 12.125 $$ - (b) For $k=12$, the roots are: $$ x = 2.5 \text{ and } x = 3 $$