Real Equal Roots 994212

1. **State the problem:** We are given the quadratic equation $$x^2 + (3k - 1)x + 10 + 2k = 0$$ and need to find: (i) The set of values of $k$ for which the equation has real roots. (ii) The values of $k$ for which the equation has equal roots and the roots themselves. 2. **Recall the discriminant condition:** For a quadratic equation $$ax^2 + bx + c = 0$$, - The roots are real if the discriminant $$\Delta = b^2 - 4ac \geq 0$$. - The roots are equal if $$\Delta = 0$$. 3. **Identify coefficients:** - $$a = 1$$ - $$b = 3k - 1$$ - $$c = 10 + 2k$$ 4. **Write the discriminant:** $$\Delta = (3k - 1)^2 - 4(1)(10 + 2k)$$ 5. **Expand and simplify:** $$\Delta = (3k - 1)^2 - 4(10 + 2k) = (3k)^2 - 2 \times 3k \times 1 + 1^2 - 40 - 8k$$ $$= 9k^2 - 6k + 1 - 40 - 8k = 9k^2 - 14k - 39$$ 6. **(i) Real roots condition:** $$9k^2 - 14k - 39 \geq 0$$ 7. **Solve the inequality:** Find roots of $$9k^2 - 14k - 39 = 0$$ using quadratic formula: $$k = \frac{14 \pm \sqrt{(-14)^2 - 4 \times 9 \times (-39)}}{2 \times 9} = \frac{14 \pm \sqrt{196 + 1404}}{18} = \frac{14 \pm \sqrt{1600}}{18}$$ $$= \frac{14 \pm 40}{18}$$ Two roots: - $$k_1 = \frac{14 - 40}{18} = \frac{-26}{18} = -\frac{13}{9}$$ - $$k_2 = \frac{14 + 40}{18} = \frac{54}{18} = 3$$ 8. **Sign analysis:** Since the leading coefficient $$9 > 0$$, the parabola opens upward. The inequality $$9k^2 - 14k - 39 \geq 0$$ holds when: $$k \leq -\frac{13}{9}$$ or $$k \geq 3$$ 9. **(ii) Equal roots condition:** Set discriminant to zero: $$9k^2 - 14k - 39 = 0$$ We already found the roots: $$k = -\frac{13}{9}$$ and $$k = 3$$ 10. **Find equal roots for each $k$:** Use quadratic formula for each $k$: - For $$k = -\frac{13}{9}$$: $$b = 3k - 1 = 3 \times \left(-\frac{13}{9}\right) - 1 = -\frac{39}{9} - 1 = -\frac{39}{9} - \frac{9}{9} = -\frac{48}{9} = -\frac{16}{3}$$ $$c = 10 + 2k = 10 + 2 \times \left(-\frac{13}{9}\right) = 10 - \frac{26}{9} = \frac{90}{9} - \frac{26}{9} = \frac{64}{9}$$ Equal root: $$x = -\frac{b}{2a} = -\frac{-\frac{16}{3}}{2} = \frac{16}{3} \times \frac{1}{2} = \frac{16}{6} = \frac{8}{3}$$ - For $$k = 3$$: $$b = 3 \times 3 - 1 = 9 - 1 = 8$$ $$c = 10 + 2 \times 3 = 10 + 6 = 16$$ Equal root: $$x = -\frac{b}{2a} = -\frac{8}{2} = -4$$ **Final answers:** (i) The equation has real roots for $$k \leq -\frac{13}{9}$$ or $$k \geq 3$$. (ii) The equation has equal roots for $$k = -\frac{13}{9}$$ and $$k = 3$$. The equal roots are: - For $$k = -\frac{13}{9}$$, root is $$x = \frac{8}{3}$$. - For $$k = 3$$, root is $$x = -4$$.