1. **Problem statement:** Find the value of $k$ for which the quadratic equation $$(2k - 1)x^2 + 4x + k = 0$$ has: A. Two real roots B. No real root C. One double root 2. **Formula used:** For a quadratic equation $ax^2 + bx + c = 0$, the discriminant $\Delta$ is given by: $$\Delta = b^2 - 4ac$$ - If $\Delta > 0$, the equation has two distinct real roots. - If $\Delta = 0$, the equation has one double root (repeated root). - If $\Delta < 0$, the equation has no real roots. 3. **Identify coefficients:** $$a = 2k - 1, \quad b = 4, \quad c = k$$ 4. **Calculate discriminant:** $$\Delta = 4^2 - 4(2k - 1)(k) = 16 - 4(2k^2 - k) = 16 - 8k^2 + 4k$$ Simplify: $$\Delta = -8k^2 + 4k + 16$$ 5. **Analyze each case:** **A. Two real roots:** $$\Delta > 0 \implies -8k^2 + 4k + 16 > 0$$ Divide entire inequality by $-4$ (note: inequality sign reverses because dividing by negative): $$2k^2 - k - 4 < 0$$ Factor or use quadratic formula for $2k^2 - k - 4 = 0$: $$k = \frac{1 \pm \sqrt{(-1)^2 - 4 \times 2 \times (-4)}}{2 \times 2} = \frac{1 \pm \sqrt{1 + 32}}{4} = \frac{1 \pm \sqrt{33}}{4}$$ Approximate roots: $$k_1 = \frac{1 - 5.7446}{4} = -1.1861, \quad k_2 = \frac{1 + 5.7446}{4} = 1.6861$$ Since $2k^2 - k - 4$ is a parabola opening upwards, it is less than zero between the roots: $$-1.1861 < k < 1.6861$$ **B. No real root:** $$\Delta < 0 \implies -8k^2 + 4k + 16 < 0$$ Similarly: $$2k^2 - k - 4 > 0$$ This holds when: $$k < -1.1861 \quad \text{or} \quad k > 1.6861$$ **C. One double root:** $$\Delta = 0 \implies -8k^2 + 4k + 16 = 0$$ Or equivalently: $$2k^2 - k - 4 = 0$$ Solutions are: $$k = \frac{1 \pm \sqrt{33}}{4}$$ 6. **Summary:** - Two real roots for $$-1.1861 < k < 1.6861$$ - No real roots for $$k < -1.1861$$ or $$k > 1.6861$$ - One double root for $$k = \frac{1 - \sqrt{33}}{4} \approx -1.1861$$ or $$k = \frac{1 + \sqrt{33}}{4} \approx 1.6861$$