Quadratic K Roots 13Ab16

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 quadratic formula and discriminant:** 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. **Calculate the discriminant:** $$\Delta = (3k - 1)^2 - 4(1)(10 + 2k)$$ $$= (3k - 1)^2 - 4(10 + 2k)$$ $$= (3k - 1)^2 - 40 - 8k$$ Expand $$(3k - 1)^2$$: $$= 9k^2 - 6k + 1$$ So, $$\Delta = 9k^2 - 6k + 1 - 40 - 8k = 9k^2 - 14k - 39$$ 5. **(i) Condition for real roots:** $$\Delta \geq 0$$ $$9k^2 - 14k - 39 \geq 0$$ 6. **Solve the inequality:** Find roots of the quadratic equation: $$9k^2 - 14k - 39 = 0$$ Use 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: $$k_1 = \frac{14 - 40}{18} = \frac{-26}{18} = -\frac{13}{9}$$ $$k_2 = \frac{14 + 40}{18} = \frac{54}{18} = 3$$ 7. **Analyze inequality $9k^2 - 14k - 39 \geq 0$:** Since the leading coefficient $9 > 0$, the parabola opens upwards. The inequality holds when: $$k \leq -\frac{13}{9}$$ or $$k \geq 3$$ 8. **(ii) Condition for equal roots:** Set discriminant to zero: $$9k^2 - 14k - 39 = 0$$ We already found the roots: $$k = -\frac{13}{9}$$ and $$k = 3$$ 9. **Find equal roots for each $k$:** Use quadratic formula: $$x = \frac{-(3k - 1)}{2}$$ since discriminant is zero. For $$k = -\frac{13}{9}$$: $$x = \frac{-(3(-\frac{13}{9}) - 1)}{2} = \frac{-( -\frac{39}{9} - 1)}{2} = \frac{-( -\frac{39}{9} - \frac{9}{9})}{2} = \frac{\frac{48}{9}}{2} = \frac{48}{18} = \frac{8}{3}$$ For $$k = 3$$: $$x = \frac{-(3(3) - 1)}{2} = \frac{-(9 - 1)}{2} = \frac{-8}{2} = -4$$ **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 $$x = -4$$.