1. **Solve the equation** $3x(x - 2)(4x + 7) = 0$. The product of factors equals zero if any factor is zero. So, solve each factor: $$3x = 0 \Rightarrow x = 0$$ $$x - 2 = 0 \Rightarrow x = 2$$ $$4x + 7 = 0 \Rightarrow x = -\frac{7}{4}$$ **Solutions:** $x = 0, 2, -\frac{7}{4}$. 2. **Solve the quadratic equation** $x^2 - 2x - 2 = 0$. Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Here, $a=1$, $b=-2$, $c=-2$. Calculate discriminant: $$\Delta = (-2)^2 - 4(1)(-2) = 4 + 8 = 12$$ So, $$x = \frac{2 \pm \sqrt{12}}{2} = \frac{2 \pm 2\sqrt{3}}{2} = 1 \pm \sqrt{3}$$ **Solutions:** $x = 1 + \sqrt{3}$ or $x = 1 - \sqrt{3}$. 3. **Solve the inequality** $x^2 - 2x - 8 < 0$ for integer $x$. Factor the quadratic: $$x^2 - 2x - 8 = (x - 4)(x + 2)$$ The inequality $(x - 4)(x + 2) < 0$ holds where the product is negative. This happens between the roots: $$-2 < x < 4$$ For integer $x$, solutions are: $$x = -1, 0, 1, 2, 3$$ 4. **Show that equation** $x^2 - 2px = -p^2$ has equal roots for all $p$. Rewrite: $$x^2 - 2px + p^2 = 0$$ This is: $$(x - p)^2 = 0$$ The discriminant: $$\Delta = (-2p)^2 - 4(1)(p^2) = 4p^2 - 4p^2 = 0$$ Zero discriminant means equal roots. 5. (a) **Solve for** $x$ in $2^x + 2^{x+1} = 192$. Factor: $$2^x + 2^{x+1} = 2^x + 2 \cdot 2^x = 3 \cdot 2^x = 192$$ So, $$2^x = \frac{192}{3} = 64 = 2^6$$ Hence, $$x = 6$$ (b) **Solve** $\sqrt{5} - x + x + 1 = 0$. Simplify: $$\sqrt{5} + 1 = 0$$ This is false since $\sqrt{5} + 1 > 0$. So, no solution. (c) **If** $\log 2 = p$ and $\log 3 = q$, find $\log_1(1/2)$. Note: $\log_1$ is undefined since base 1 is invalid. Assuming typo, if it means $\log_{10}(1/2)$: $$\log(1/2) = \log 1 - \log 2 = 0 - p = -p$$ (d) **Show that if** $\log x + \log z = \log y^2$, then $x, y, z$ form a geometric sequence. Use log properties: $$\log x + \log z = \log (xz) = \log y^2$$ So, $$xz = y^2$$ This implies: $$y = \sqrt{xz}$$ Hence, $y$ is the geometric mean of $x$ and $z$, so $x, y, z$ are in geometric progression. 6. **Derivative of** $f(x) = -2x^2 - 2$. $$f'(x) = -4x$$ 7. **Given** $f(x) = p^x$ and point $A(2, 25/9)$ lies on $f$. Find $p$: $$p^2 = \frac{25}{9}$$ $$p = \frac{5}{3}$$ 8. **Equation of** $g(x) = x^3 - 3x - 2$. (a) Find coordinates of $B$ (y-intercept): $$B = (0, g(0)) = (0, -2)$$ (b) Find turning points by solving $g'(x) = 0$: $$g'(x) = 3x^2 - 3 = 0 \Rightarrow x^2 = 1 \Rightarrow x = \pm 1$$ Find $g(1)$ and $g(-1)$: $$g(1) = 1 - 3 - 2 = -4$$ $$g(-1) = -1 + 3 - 2 = 0$$ So turning points at $(1, -4)$ and $(-1, 0)$. 9. **Minimum surface area of cylinder with fixed volume** $V = 150\pi$. Given: $$S = 2\pi r^2 + 2\pi r h$$ $$V = \pi r^2 h = 150\pi \Rightarrow h = \frac{150}{r^2}$$ Substitute $h$ into $S$: $$S = 2\pi r^2 + 2\pi r \cdot \frac{150}{r^2} = 2\pi r^2 + \frac{300\pi}{r}$$ Differentiate $S$ w.r.t. $r$: $$\frac{dS}{dr} = 4\pi r - \frac{300\pi}{r^2}$$ Set derivative to zero for minimum: $$4\pi r = \frac{300\pi}{r^2} \Rightarrow 4r^3 = 300 \Rightarrow r^3 = 75$$ $$r = \sqrt[3]{75}$$ This radius minimizes surface area. **Final answers:** - Equation roots: $x=0, 2, -\frac{7}{4}$ - Quadratic roots: $x=1 \pm \sqrt{3}$ - Inequality integer solutions: $x = -1, 0, 1, 2, 3$ - Equal roots for $x^2 - 2px = -p^2$ proven. - $x=6$ solves $2^x + 2^{x+1} = 192$. - No solution for $\sqrt{5} - x + x + 1 = 0$. - $\log(1/2) = -p$. - $x,y,z$ form geometric sequence if $\log x + \log z = \log y^2$. - $f'(x) = -4x$. - $p = \frac{5}{3}$. - $B = (0, -2)$, turning points at $(1, -4)$ and $(-1, 0)$. - Cylinder radius minimizing surface area: $r = \sqrt[3]{75}$.