1. **State the problem:** We have the quadratic equation $$x^2 + kx + 15 - k = 0$$ and need to find the values of $$k$$ such that it has two distinct real roots. 2. **Formula and rules:** For a quadratic equation $$ax^2 + bx + c = 0$$, the discriminant $$\Delta = b^2 - 4ac$$ determines the nature of roots. - Two distinct real roots exist if and only if $$\Delta > 0$$. 3. **Apply to our equation:** Here, $$a=1$$, $$b=k$$, and $$c=15-k$$. Calculate discriminant: $$\Delta = k^2 - 4(1)(15 - k) = k^2 - 60 + 4k = k^2 + 4k - 60$$ 4. **Condition for two distinct real roots:** $$k^2 + 4k - 60 > 0$$ 5. **Solve inequality:** Find roots of quadratic equation $$k^2 + 4k - 60 = 0$$: $$k = \frac{-4 \pm \sqrt{4^2 - 4 \times 1 \times (-60)}}{2} = \frac{-4 \pm \sqrt{16 + 240}}{2} = \frac{-4 \pm \sqrt{256}}{2} = \frac{-4 \pm 16}{2}$$ So roots are: $$k = \frac{-4 + 16}{2} = 6$$ and $$k = \frac{-4 - 16}{2} = -10$$ 6. **Sign analysis:** Since the parabola opens upward (coefficient of $$k^2$$ is positive), the inequality $$k^2 + 4k - 60 > 0$$ holds when $$k < -10$$ or $$k > 6$$. **Answer for (a):** $$k < -10$$ or $$k > 6$$. --- 7. **Part (b):** Find $$k$$ such that the two distinct real roots are both positive or both negative. 8. **Let roots be $$r_1$$ and $$r_2$$.** From Vieta's formulas: $$r_1 + r_2 = -k$$ $$r_1 r_2 = 15 - k$$ 9. **Both roots positive:** - Sum of roots $$r_1 + r_2 > 0$$ - Product of roots $$r_1 r_2 > 0$$ So: $$-k > 0 \implies k < 0$$ $$15 - k > 0 \implies k < 15$$ Since $$k < 0$$ is more restrictive, for both roots positive, $$k < 0$$. 10. **Both roots negative:** - Sum of roots $$r_1 + r_2 < 0$$ - Product of roots $$r_1 r_2 > 0$$ So: $$-k < 0 \implies k > 0$$ $$15 - k > 0 \implies k < 15$$ So for both roots negative, $$0 < k < 15$$. 11. **Combine with distinct roots condition:** From (a), $$k < -10$$ or $$k > 6$$. - For both roots positive and distinct: $$k < -10$$ (since $$k < 0$$ and $$k < -10$$ overlap only at $$k < -10$$). - For both roots negative and distinct: $$6 < k < 15$$ (since $$k > 6$$ and $$0 < k < 15$$ overlap at $$6 < k < 15$$). **Answer for (b):** $$k < -10$$ or $$6 < k < 15$$.