1. **State the problem:** We have the quadratic equation $$k + 9 = (6 - x)(2x + 1)$$ and need to find: (a) The range of values of $$k$$ for which the equation has two distinct real roots. (b) Solve the quadratic equation when $$k$$ is the largest possible integer in that range. 2. **Rewrite the equation:** Expand the right side: $$(6 - x)(2x + 1) = 6 \times 2x + 6 \times 1 - x \times 2x - x \times 1 = 12x + 6 - 2x^2 - x = -2x^2 + 11x + 6$$ So the equation becomes: $$k + 9 = -2x^2 + 11x + 6$$ Rearranged to standard quadratic form: $$-2x^2 + 11x + 6 - k - 9 = 0 \implies -2x^2 + 11x - (k + 3) = 0$$ 3. **Identify coefficients:** $$a = -2, \quad b = 11, \quad c = -(k + 3)$$ 4. **Condition for two distinct real roots:** The discriminant $$\Delta$$ must be positive: $$\Delta = b^2 - 4ac > 0$$ Substitute values: $$11^2 - 4(-2)(-(k + 3)) > 0$$ $$121 - 8(k + 3) > 0$$ 5. **Solve inequality:** $$121 - 8k - 24 > 0$$ $$97 - 8k > 0$$ $$-8k > -97$$ Divide both sides by -8 (remember to reverse inequality): $$\cancel{-8}k > \cancel{-97}$$ $$k < \frac{97}{8}$$ 6. **Range of $$k$$:** $$k < 12.125$$ 7. **Find largest integer $$k$$:** Largest integer less than 12.125 is $$k = 12$$. 8. **Solve quadratic for $$k=12$$:** Equation: $$-2x^2 + 11x - (12 + 3) = 0 \implies -2x^2 + 11x - 15 = 0$$ Multiply both sides by $$-1$$ for simplicity: $$2x^2 - 11x + 15 = 0$$ 9. **Use quadratic formula:** $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $$a=2$$, $$b=-11$$, $$c=15$$. Calculate discriminant: $$\Delta = (-11)^2 - 4 \times 2 \times 15 = 121 - 120 = 1$$ 10. **Calculate roots:** $$x = \frac{-(-11) \pm \sqrt{1}}{2 \times 2} = \frac{11 \pm 1}{4}$$ 11. **Final roots:** $$x_1 = \frac{11 + 1}{4} = \frac{12}{4} = 3$$ $$x_2 = \frac{11 - 1}{4} = \frac{10}{4} = 2.5$$ **Answer:** (a) The range of $$k$$ for two distinct real roots is $$k < 12.125$$. (b) For $$k=12$$, the roots are $$x=3$$ and $$x=2.5$$.