1. **Problem 4:** Find values of $k$ such that the equation $$p(x+1) + q(x-r) = \frac{k}{2}x$$ has two equal roots. 2. **Rewrite the equation:** $$p(x+1) + q(x-r) = \frac{k}{2}x$$ Expanding the left side: $$px + p + qx - qr = \frac{k}{2}x$$ Combine like terms: $$(p + q)x + (p - qr) = \frac{k}{2}x$$ 3. **Bring all terms to one side:** $$(p + q)x + (p - qr) - \frac{k}{2}x = 0$$ Combine $x$ terms: $$\left(p + q - \frac{k}{2}\right)x + (p - qr) = 0$$ 4. **This is a linear equation in $x$, not quadratic, so it cannot have two equal roots unless it is identically zero or degenerate.** **Assuming the problem means the quadratic formed by rearranging or interpreting differently, let's consider the quadratic form:** Rewrite as: $$p(x+1) + q(x-r) - \frac{k}{2}x = 0$$ If we multiply both sides by 2 to clear denominator: $$2p(x+1) + 2q(x-r) - kx = 0$$ Expanding: $$2px + 2p + 2qx - 2qr - kx = 0$$ Group $x$ terms: $$(2p + 2q - k)x + (2p - 2qr) = 0$$ Again, linear in $x$. For two equal roots, the equation must be quadratic. So likely the problem is misstated or missing context. **If the problem is to find $k$ such that the quadratic equation formed by $p(x+1) + q(x-r) = \frac{k}{2}x$ has two equal roots, then the quadratic must be formed by rearranging terms or squaring.** **Without additional info, we cannot proceed further on problem 4.** --- 5. **Problem 5:** For what value of $k$ do the quadratic equations $$6x^2 - 17x + 12 = 0$$ and $$3x^2 - 2x + k = 0$$ have a common root? 6. **Let the common root be $\alpha$. Then $\alpha$ satisfies both equations:** $$6\alpha^2 - 17\alpha + 12 = 0$$ $$3\alpha^2 - 2\alpha + k = 0$$ 7. **Express $k$ in terms of $\alpha$ from the second equation:** $$k = -3\alpha^2 + 2\alpha$$ 8. **Since $\alpha$ satisfies the first equation, we can use it to eliminate $\alpha^2$:** From first equation: $$6\alpha^2 = 17\alpha - 12$$ Divide both sides by 6: $$\alpha^2 = \frac{17\alpha - 12}{6}$$ 9. **Substitute $\alpha^2$ into expression for $k$:** $$k = -3 \times \frac{17\alpha - 12}{6} + 2\alpha = -\frac{3}{6}(17\alpha - 12) + 2\alpha = -\frac{1}{2}(17\alpha - 12) + 2\alpha$$ Simplify: $$k = -\frac{17}{2}\alpha + 6 + 2\alpha = \left(-\frac{17}{2} + 2\right)\alpha + 6 = -\frac{13}{2}\alpha + 6$$ 10. **To find $k$, we need $\alpha$. Since $\alpha$ is a root of the first quadratic, solve it:** $$6x^2 - 17x + 12 = 0$$ Discriminant: $$\Delta = (-17)^2 - 4 \times 6 \times 12 = 289 - 288 = 1$$ Roots: $$x = \frac{17 \pm 1}{2 \times 6} = \frac{17 \pm 1}{12}$$ So roots are: $$x_1 = \frac{18}{12} = \frac{3}{2} = 1.5$$ $$x_2 = \frac{16}{12} = \frac{4}{3} \approx 1.333$$ 11. **Calculate $k$ for each root:** - For $\alpha = \frac{3}{2}$: $$k = -\frac{13}{2} \times \frac{3}{2} + 6 = -\frac{39}{4} + 6 = -9.75 + 6 = -3.75$$ - For $\alpha = \frac{4}{3}$: $$k = -\frac{13}{2} \times \frac{4}{3} + 6 = -\frac{52}{6} + 6 = -8.6667 + 6 = -2.6667$$ 12. **Final answer:** $$k = -\frac{15}{4} \quad \text{or} \quad k = -\frac{8}{3}$$ --- **Summary:** - Problem 4: Insufficient information or equation is linear, no quadratic roots. - Problem 5: Values of $k$ for common root are $$k = -\frac{15}{4}$$ or $$k = -\frac{8}{3}$$.