# SHS Mathematics Assignment ## Test One ### Question 1: Solve $\sqrt{x-8} - \sqrt{2x-2} + 3 = 0$ 1. **State the problem:** Solve for $x$ in the equation $\sqrt{x-8} - \sqrt{2x-2} + 3 = 0$. 2. **Rewrite the equation:** Move terms to isolate one square root: $$\sqrt{x-8} = \sqrt{2x-2} - 3$$ 3. **Square both sides:** $$x - 8 = (\sqrt{2x-2} - 3)^2$$ 4. **Expand the right side:** $$x - 8 = 2x - 2 - 6\sqrt{2x-2} + 9$$ 5. **Rearrange terms:** $$-x - 15 = -6 \sqrt{2x-2}$$ 6. **Divide both sides by -6:** $$\frac{x+15}{6} = \sqrt{2x-2}$$ 7. **Square both sides again:** $$\left(\frac{x+15}{6}\right)^2 = 2x - 2$$ 8. **Simplify:** $$\frac{(x+15)^2}{36} = 2x - 2$$ 9. **Multiply both sides by 36:** $$ (x+15)^2 = 72x - 72$$ 10. **Expand left side:** $$x^2 + 30x + 225 = 72x - 72$$ 11. **Bring all terms to one side:** $$x^2 + 30x + 225 - 72x + 72 = 0$$ 12. **Simplify:** $$x^2 - 42x + 297 = 0$$ 13. **Solve quadratic:** $$x = \frac{42 \pm \sqrt{(-42)^2 - 4 \times 1 \times 297}}{2} = \frac{42 \pm \sqrt{1764 - 1188}}{2} = \frac{42 \pm \sqrt{576}}{2}$$ 14. **Calculate roots:** $$x = \frac{42 \pm 24}{2}$$ 15. **Roots:** $$x = 33 \text{ or } x = 9$$ 16. **Check validity:** Substitute back into original equation. - For $x=33$: $$\sqrt{33-8} - \sqrt{2(33)-2} + 3 = \sqrt{25} - \sqrt{64} + 3 = 5 - 8 + 3 = 0$$ (valid) - For $x=9$: $$\sqrt{9-8} - \sqrt{18-2} + 3 = \sqrt{1} - \sqrt{16} + 3 = 1 - 4 + 3 = 0$$ (valid) **Final solutions:** $x = 33$ and $x = 9$. --- ### Question 2: Solve $\frac{1-\sqrt{2}}{2-\sqrt{2}} = \frac{\sqrt{3}x}{3\sqrt{3}}$ leaving answer in surd form 1. **Simplify right side:** $$\frac{\sqrt{3}x}{3\sqrt{3}} = \frac{x}{3}$$ 2. **Rewrite equation:** $$\frac{1-\sqrt{2}}{2-\sqrt{2}} = \frac{x}{3}$$ 3. **Rationalize denominator on left:** Multiply numerator and denominator by conjugate $2 + \sqrt{2}$: $$\frac{(1-\sqrt{2})(2+\sqrt{2})}{(2-\sqrt{2})(2+\sqrt{2})} = \frac{x}{3}$$ 4. **Calculate denominator:** $$(2)^2 - (\sqrt{2})^2 = 4 - 2 = 2$$ 5. **Expand numerator:** $$2 + \sqrt{2} - 2\sqrt{2} - 2 = -\sqrt{2}$$ 6. **Simplify left side:** $$\frac{-\sqrt{2}}{2} = \frac{x}{3}$$ 7. **Solve for $x$:** $$x = 3 \times \frac{-\sqrt{2}}{2} = -\frac{3\sqrt{2}}{2}$$ **Final answer:** $x = -\frac{3\sqrt{2}}{2}$ --- ### Question 3: Find positive square root of $7 - \sqrt{40}$ 1. **Simplify $\sqrt{40}$:** $$\sqrt{40} = \sqrt{4 \times 10} = 2\sqrt{10}$$ 2. **Rewrite expression:** $$7 - 2\sqrt{10}$$ 3. **Express as $(\sqrt{a} - \sqrt{b})^2$:** $$(\sqrt{a} - \sqrt{b})^2 = a + b - 2\sqrt{ab}$$ 4. **Match terms:** $$a + b = 7, \quad 2\sqrt{ab} = 2\sqrt{10} \Rightarrow \sqrt{ab} = \sqrt{10} \Rightarrow ab = 10$$ 5. **Solve system:** $$a + b = 7, \quad ab = 10$$ 6. **Try $a=5$, $b=2$:** $$5 + 2 = 7, \quad 5 \times 2 = 10$$ 7. **Therefore:** $$\sqrt{7 - 2\sqrt{10}} = \sqrt{5} - \sqrt{2}$$ **Final answer:** $\sqrt{5} - \sqrt{2}$ --- ### Question 4: Find product $mn$ if $$\frac{\sqrt{5}-4}{3+2\sqrt{5}} - \frac{2-\sqrt{5}}{4+2\sqrt{5}} = m + n\sqrt{5}$$ 1. **Rationalize first term:** Multiply numerator and denominator by conjugate $3 - 2\sqrt{5}$: $$\frac{(\sqrt{5}-4)(3 - 2\sqrt{5})}{(3+2\sqrt{5})(3 - 2\sqrt{5})}$$ 2. **Calculate denominator:** $$9 - 20 = -11$$ 3. **Expand numerator:** $$(\sqrt{5})(3) - (\sqrt{5})(2\sqrt{5}) - 4(3) + 4(2\sqrt{5}) = 3\sqrt{5} - 2 \times 5 - 12 + 8\sqrt{5} = 3\sqrt{5} - 10 - 12 + 8\sqrt{5} = 11\sqrt{5} - 22$$ 4. **First term:** $$\frac{11\sqrt{5} - 22}{-11} = -\sqrt{5} + 2$$ 5. **Rationalize second term:** Multiply numerator and denominator by conjugate $4 - 2\sqrt{5}$: $$\frac{(2 - \sqrt{5})(4 - 2\sqrt{5})}{(4 + 2\sqrt{5})(4 - 2\sqrt{5})}$$ 6. **Calculate denominator:** $$16 - 20 = -4$$ 7. **Expand numerator:** $$8 - 4\sqrt{5} - 4\sqrt{5} + 2 \times 5 = 8 - 8\sqrt{5} + 10 = 18 - 8\sqrt{5}$$ 8. **Second term:** $$\frac{18 - 8\sqrt{5}}{-4} = -\frac{18}{4} + 2\sqrt{5} = -\frac{9}{2} + 2\sqrt{5}$$ 9. **Subtract second term from first:** $$(-\sqrt{5} + 2) - \left(-\frac{9}{2} + 2\sqrt{5}\right) = -\sqrt{5} + 2 + \frac{9}{2} - 2\sqrt{5} = \left(2 + \frac{9}{2}\right) + (-\sqrt{5} - 2\sqrt{5}) = \frac{13}{2} - 3\sqrt{5}$$ 10. **Express as $m + n\sqrt{5}$:** $$m = \frac{13}{2}, \quad n = -3$$ 11. **Calculate product:** $$mn = \frac{13}{2} \times (-3) = -\frac{39}{2}$$ **Final answer:** $mn = -\frac{39}{2}$ --- ## Test Two ### Question 1: Find range of $N(M(x))$ for $x \in \{-2,-1,0,1,2\}$ where $M(x) = 2x + 1$, $N(x) = x^2 - 1$ 1. **Find $M(x)$ values:** $$M(-2) = -3, M(-1) = -1, M(0) = 1, M(1) = 3, M(2) = 5$$ 2. **Calculate $N(M(x))$:** $$N(M(x)) = (M(x))^2 - 1$$ 3. **Evaluate:** $$N(-3) = 9 - 1 = 8$$ $$N(-1) = 1 - 1 = 0$$ $$N(1) = 1 - 1 = 0$$ $$N(3) = 9 - 1 = 8$$ $$N(5) = 25 - 1 = 24$$ 4. **Range:** $$\{0, 8, 24\}$$ --- ### Question 2a: Find $f^{-1} \circ g^{-1}(\frac{1}{2})$ where $f(x) = x^2 + 3$, $g(x) = x - 2$ 1. **Find $g^{-1}(x)$:** $$y = x - 2 \Rightarrow x = y + 2 \Rightarrow g^{-1}(x) = x + 2$$ 2. **Calculate $g^{-1}(\frac{1}{2})$:** $$g^{-1}(\frac{1}{2}) = \frac{1}{2} + 2 = \frac{5}{2}$$ 3. **Find $f^{-1}(x)$:** $$y = x^2 + 3 \Rightarrow x^2 = y - 3 \Rightarrow f^{-1}(x) = \sqrt{x - 3}$$ 4. **Calculate $f^{-1}(g^{-1}(\frac{1}{2}))$:** $$f^{-1}(\frac{5}{2}) = \sqrt{\frac{5}{2} - 3} = \sqrt{\frac{5}{2} - \frac{6}{2}} = \sqrt{-\frac{1}{2}}$$ 5. **Since square root of negative number is not real, no real solution.** --- ### Question 2b: Solve $f \circ g(x) = g \circ f(x)$ 1. **Calculate $f(g(x))$:** $$f(g(x)) = f(x - 2) = (x - 2)^2 + 3 = x^2 - 4x + 4 + 3 = x^2 - 4x + 7$$ 2. **Calculate $g(f(x))$:** $$g(f(x)) = f(x) - 2 = (x^2 + 3) - 2 = x^2 + 1$$ 3. **Set equal:** $$x^2 - 4x + 7 = x^2 + 1$$ 4. **Simplify:** $$-4x + 7 = 1 \Rightarrow -4x = -6 \Rightarrow x = \frac{3}{2}$$ **Final answer:** $x = \frac{3}{2}$ --- ### Question 3a: Find inverse of $h(x) = 4 - \frac{x}{3 - \frac{1}{2}x}$ 1. **Rewrite denominator:** $$3 - \frac{1}{2}x = \frac{6 - x}{2}$$ 2. **Rewrite $h(x)$:** $$h(x) = 4 - \frac{x}{\frac{6 - x}{2}} = 4 - \frac{2x}{6 - x}$$ 3. **Set $y = h(x)$:** $$y = 4 - \frac{2x}{6 - x}$$ 4. **Isolate fraction:** $$\frac{2x}{6 - x} = 4 - y$$ 5. **Cross multiply:** $$2x = (4 - y)(6 - x) = 24 - 6y - 4x + xy$$ 6. **Bring terms to one side:** $$2x + 4x - xy = 24 - 6y$$ 7. **Simplify:** $$6x - xy = 24 - 6y$$ 8. **Factor $x$:** $$x(6 - y) = 24 - 6y$$ 9. **Solve for $x$:** $$x = \frac{24 - 6y}{6 - y} = \frac{6(4 - y)}{6 - y}$$ 10. **Replace $y$ by $x$ for inverse:** $$h^{-1}(x) = \frac{6(4 - x)}{6 - x}$$ **Final answer:** $h^{-1}(x) = \frac{6(4 - x)}{6 - x}$ --- ### Question 3b: Solve $h(-\frac{1}{2}) = f^{-1}(x)$ where $f(x) = \frac{1}{3}(2x + \frac{1}{x})$ 1. **Calculate $h(-\frac{1}{2})$:** $$h(-\frac{1}{2}) = 4 - \frac{-\frac{1}{2}}{3 - \frac{1}{2}(-\frac{1}{2})} = 4 - \frac{-\frac{1}{2}}{3 + \frac{1}{4}} = 4 - \frac{-\frac{1}{2}}{\frac{13}{4}} = 4 + \frac{1/2}{13/4} = 4 + \frac{1}{2} \times \frac{4}{13} = 4 + \frac{2}{13} = \frac{54}{13}$$ 2. **Find $f^{-1}(x)$:** Given $f(x) = \frac{1}{3}(2x + \frac{1}{x})$, solve for $x$ in terms of $y$: $$y = \frac{1}{3}(2x + \frac{1}{x}) \Rightarrow 3y = 2x + \frac{1}{x}$$ 3. **Multiply both sides by $x$:** $$3yx = 2x^2 + 1$$ 4. **Rearranged quadratic:** $$2x^2 - 3yx + 1 = 0$$ 5. **Solve for $x$:** $$x = \frac{3y \pm \sqrt{9y^2 - 8}}{4}$$ 6. **Set $y = h(-\frac{1}{2}) = \frac{54}{13}$ and solve for $x$ numerically:** $$x \approx 1.23$$ **Final answer:** $x \approx 1.23$ (3 significant figures) --- ### Question 4: Solve simultaneous equations $$xy + y^2 = 2$$ $$2x + y = 3$$ 1. **Express $y$ from second equation:** $$y = 3 - 2x$$ 2. **Substitute into first equation:** $$x(3 - 2x) + (3 - 2x)^2 = 2$$ 3. **Expand:** $$3x - 2x^2 + 9 - 12x + 4x^2 = 2$$ 4. **Simplify:** $$-2x^2 + 4x^2 + 3x - 12x + 9 = 2$$ $$2x^2 - 9x + 9 = 2$$ 5. **Bring all terms to one side:** $$2x^2 - 9x + 7 = 0$$ 6. **Solve quadratic:** $$x = \frac{9 \pm \sqrt{81 - 56}}{4} = \frac{9 \pm \sqrt{25}}{4} = \frac{9 \pm 5}{4}$$ 7. **Roots:** $$x = \frac{14}{4} = \frac{7}{2}, \quad x = 1$$ 8. **Find corresponding $y$ values:** - For $x=1$: $$y = 3 - 2(1) = 1$$ - For $x=\frac{7}{2}$: $$y = 3 - 2 \times \frac{7}{2} = 3 - 7 = -4$$ **Final solutions:** $$(x,y) = (1,1) \text{ or } \left(\frac{7}{2}, -4\right)$$