1. **Stating the problem:** We have the quadratic equation $$2x^2 + kx + 5k = 0$$ and multiple choice options for the value of $k$. To solve for $k$, we use the fact that the quadratic has real roots or find the value of $k$ satisfying a given root condition. 2. **Considering each root option:** The options given appear to be possible roots for $x$: $$-\frac{3}{2}, -\frac{2}{3}, 3, \frac{3}{2}, \frac{2}{3}$$. 3. **Substitute each root into the quadratic equation:** For each root $x=r$, plug into the equation and solve for $k$ so the equation holds true. 4. **Example with $x = -\frac{3}{2}$:** $$2\left(-\frac{3}{2}\right)^2 + k\left(-\frac{3}{2}\right) + 5k = 0$$ $$2 \cdot \frac{9}{4} - \frac{3}{2}k + 5k = 0$$ $$\frac{9}{2} + \frac{7}{2}k = 0$$ $$\frac{7}{2}k = -\frac{9}{2}$$ $$k = -\frac{9}{7}$$ this does not match any option given. 5. **Repeat process for $x = -\frac{2}{3}$:** $$2\left(-\frac{2}{3}\right)^2 + k\left(-\frac{2}{3}\right) + 5k = 0$$ $$2 \cdot \frac{4}{9} - \frac{2}{3}k + 5k = 0$$ $$\frac{8}{9} + \frac{13}{3}k = 0$$ $$\frac{13}{3}k = -\frac{8}{9}$$ $$k = -\frac{8}{9} \cdot \frac{3}{13} = -\frac{8}{39}$$, not matching options. 6. **Try $x = 3$:** $$2(3)^2 + 3k + 5k = 0$$ $$2 \cdot 9 + 8k = 0$$ $$18 + 8k = 0$$ $$8k = -18$$ $$k = -\frac{9}{4}$$, not listed. 7. **Try $x = \frac{3}{2}$:** $$2\left(\frac{3}{2}\right)^2 + k\left(\frac{3}{2}\right) + 5k = 0$$ $$2 \cdot \frac{9}{4} + \frac{3}{2}k + 5k = 0$$ $$\frac{9}{2} + \frac{13}{2}k = 0$$ $$\frac{13}{2}k = -\frac{9}{2}$$ $$k = -\frac{9}{13}$$, not matching options. 8. **Try $x = \frac{2}{3}$:** $$2\left(\frac{2}{3}\right)^2 + k\left(\frac{2}{3}\right) + 5k = 0$$ $$2 \cdot \frac{4}{9} + \frac{2}{3}k + 5k = 0$$ $$\frac{8}{9} + \frac{17}{3}k = 0$$ $$\frac{17}{3}k = -\frac{8}{9}$$ $$k = -\frac{8}{9} \cdot \frac{3}{17} = -\frac{8}{51}$$, no match. 9. **Verify the quadratic has roots $x = -\frac{3}{2}$ and $x = -2/3$.** 10. Alternatively, the multiple choices could be for $k$, and we check which value of $k$ allows a root with the quadratic equal zero. 11. Try checking the discriminant $$\Delta = b^2 - 4ac = k^2 - 4 \, \cdot \, 2 \, \cdot \, 5k = k^2 - 40k$$ 12. For the quadratic to have real roots, the discriminant must be non-negative: $$k^2 - 40k \geq 0$$ $$k(k-40) \geq 0$$ So, either $$k \leq 0$$ or $$k \geq 40$$. 13. So negative or large positive values satisfy this. 14. Among given options: $$-\frac{3}{2} = -1.5$$, $$-\frac{2}{3} = -0.666...$$, $$3$$, $$\frac{3}{2} = 1.5$$, $$\frac{2}{3} = 0.666...$$. Only negatives satisfy $k \leq 0$; check if they satisfy the original equation. 15. Let's test each value in the original equation: for $k=-\frac{3}{2}$ Original: $$2x^2 + kx + 5k = 0$$ Substitute $k = -\frac{3}{2}$ $$2x^2 - \frac{3}{2} x - \frac{15}{2} = 0$$ Multiply both sides by 2: $$4x^2 - 3x - 15 = 0$$ Calculate discriminant: $$\Delta= (-3)^2 - 4(4)(-15) = 9 + 240 = 249 > 0$$, so roots are real. The original problem likely asks for $k$ values making the equation have real roots or matching the options. **Final Answer: Among the given options,** $k = -\frac{3}{2}$ **makes the quadratic have real roots.**