Use the formula $\sqrt{ab} = \sqrt{a} \times \sqrt{b}$ for the entire question.

(a)

\begin{align} \sqrt{18} & = \sqrt{9}\times \sqrt{2} \\ & = 3\sqrt{2} \end{align}

(b)

\begin{align} \sqrt{300} & = \sqrt{100} \times \sqrt{3} \\ & = 10\sqrt{3} \end{align}

(c)

\begin{align} \sqrt{80} & = \sqrt{16} \times \sqrt{5} \\ & = 4\sqrt{5} \end{align}

(d)

\begin{align} \sqrt{48} & = \sqrt{16} \times \sqrt{3} \\ & = 4\sqrt{3} \end{align}

(a)

\begin{align} 2\sqrt{3} + 5\sqrt{3} - 3\sqrt{3} & = \sqrt{3} (2 + 5 -3) [\phantom{000} \text{Factorise}] \\ & = 4\sqrt{3} \end{align}

(b)

\begin{align} 4\sqrt{2} - \sqrt{8} + \sqrt{50} & = 4\sqrt{2} - \sqrt{4}\times\sqrt{2} + \sqrt{25} \times \sqrt{2} \\ & = 4\sqrt{2} - 2\sqrt{2} + 5\sqrt{2} \\ & = \sqrt{2}(4 - 2 + 5) \\ & = 7\sqrt{2} \end{align}

(a) Note: I used the identity (a + b)(a - b) = a² - b²

\begin{align} (2 + \sqrt{5})(2 - \sqrt{5}) & = (2)^2 - (\sqrt{5})^2 \\ & = 4 - 5 \\ & = -1 \end{align}

(b)

\begin{align} (1 + \sqrt{3})(2 - \sqrt{3}) & = 2 - \sqrt{3} + 2\sqrt{3} - 3 \\ & = \sqrt{3} - 1 \\ & = -1 + \sqrt{3} \end{align}

(c) Note: I used the identity (a + b)(a - b) = a² - b²

\begin{align} (1 - 2\sqrt{7})^2 & = (1)^2 - 2(1)(2\sqrt{7}) + (2\sqrt{7})^2 \\ & = 1 - 4\sqrt{7} + (4)(7) \\ & = 29 - 4\sqrt{7} \end{align}

(d) Use the identity (a + b)² = a² + 2ab + b²

\begin{align} (5 + \sqrt{18})^2 & = (5)^2 + 2(5)(\sqrt{18}) + (\sqrt{18})^2 \\ & = 25 + 10\sqrt{18} + 18 \\ & = 43 + 10\sqrt{18} \\ & = 43 + 10\sqrt{9 \times 2} \\ & = 43 + 10(\sqrt{9} \times \sqrt{2}) \\ & = 43 + 10(3\sqrt{2}) \\ & = 43 + 30\sqrt{2} \end{align}

Note: I used the identities:

• (a + b)² = a² + 2ab + b²
• (a - b)² = a² - 2ab + b²

\begin{align} (7 + 3\sqrt{2})^2 - (5 - 2\sqrt{5})^2 & = [(7)^2 + 2(7)(3\sqrt{2}) + (3\sqrt{2})^2] - [(5)^2 - 2(5)(2\sqrt{5}) + (2\sqrt{5})^2] \\ & = (49 + 42\sqrt{2} + 18) - (25 - 20\sqrt{5} + 20) \\ & = (67 + 42\sqrt{2}) - (45 - 20\sqrt{5}) \\ & = 67 + 42\sqrt{2} - 45 + 20\sqrt{5} \\ & = 22 + 42\sqrt{2} + 20\sqrt{5} \end{align}

(a)

\begin{align} {6 \over \sqrt{3}} & = {6 \over \sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} \\ & = {6\sqrt{3} \over 3} \\ & = 2\sqrt{3} \end{align}

(b)

\begin{align} {\sqrt{2} \over 2\sqrt{3}} & = {\sqrt{2} \over 2\sqrt{3}} \times {2\sqrt{3} \over 2\sqrt{3}} \\ & = {2\sqrt{6} \over 12} \\ & = {\sqrt{6} \over 6} \end{align}

(c)

\begin{align} {12 \over 2\sqrt{5} - 4} & = {12 \over 2\sqrt{5} - 4} \times {2\sqrt{5} + 4 \over 2\sqrt{5} + 4} \\ & = {12(2\sqrt{5} + 4) \over (2\sqrt{5} - 4)(2\sqrt{5} + 4)} \\ & = {24\sqrt{5} + 48 \over [(2\sqrt{5})^2 - (4)^2]} \\ & = {24\sqrt{5} + 48 \over 20 - 16} \\ & = {24\sqrt{5} + 48 \over 4} \\ & = {24\sqrt{5} \over 4} + {48 \over 4} \\ & = 6\sqrt{5} + 12 \end{align}

(a)

\begin{align} {3\sqrt{2} - 4 \over 4 + 3\sqrt{2}} & = {3\sqrt{2} - 4 \over 4 + 3\sqrt{2}} \times {4 - 3\sqrt{2} \over 4 - 3\sqrt{2}} \\ & = {(3\sqrt{2} - 4)(4 - 3\sqrt{2}) \over (4 + 3\sqrt{2})(4 - 3\sqrt{2})} \\ & = {12\sqrt{2} - (3\sqrt{2})^2 - 16 + 12\sqrt{2} \over 4^2 - (3\sqrt{2})^2} \\ & = {12\sqrt{2} - 18 - 16 + 12\sqrt{2} \over 16 - 18} \\ & = {24\sqrt{2} - 34 \over -2} \\ & = {24\sqrt{2} \over -2} - {34 \over -2} \\ & = -12\sqrt{2} - (-17) \\ & = -12\sqrt{2} + 17 \end{align}

(b)

\begin{align} {\sqrt{3} + 2\sqrt{2} \over \sqrt{3} - 2\sqrt{2}} & = {\sqrt{3} + 2\sqrt{2} \over \sqrt{3} - 2\sqrt{2}} \times {\sqrt{3} + 2\sqrt{2} \over \sqrt{3} + 2\sqrt{2}} \\ & = {(\sqrt{3} + 2\sqrt{2})^2 \over (\sqrt{3} - 2\sqrt{2})(\sqrt{3} + 2\sqrt{2}) } \\ & = {(\sqrt{3})^2 + 2(\sqrt{3})(2\sqrt{2}) + (2\sqrt{2})^2 \over (\sqrt{3})^2 - (2\sqrt{2})^2} \\ & = {3 + 4\sqrt{6} + 8 \over 3 - 8} \\ & = {11 + 4\sqrt{6} \over -5} \\ & = -{11 + 4\sqrt{6} \over 5} \\ & = {-(11 + 4\sqrt{6}) \over 5} \\ & = {-11 - 4\sqrt{6} \over 5} \end{align}

(c)

\begin{align} {8 \over (\sqrt{6} + 2)^2} & = {8 \over (\sqrt{6})^2 + 2(\sqrt{6})(2) + (2)^2 } \\ & = {8 \over 6 + 4\sqrt{6} + 4} \\ & = {8 \over 10 + 4\sqrt{6}} \times {10 - 4\sqrt{6} \over 10 - 4\sqrt{6}} \phantom{00000} [\text{Rationalise denominator}] \\ & = {8(10 - 4\sqrt{6}) \over (10 + 4\sqrt{6})(10 - 4\sqrt{6})} \\ & = {80 - 32\sqrt{6} \over (10)^2 - (4\sqrt{6})^2} \\ & = {80 - 32\sqrt{6} \over 100 - 96} \\ & = {80 - 32\sqrt{6} \over 4} \\ & = {80 \over 4} - {32 \sqrt{6} \over 4} \\ & = 20 - 8\sqrt{6} \end{align}

(d)

\begin{align} {\sqrt{18} - \sqrt{12} \over \sqrt{50} - \sqrt{48}} & = {\sqrt{9} \times \sqrt{2} - \sqrt{4} \times \sqrt{3} \over \sqrt{25} \times \sqrt{2} - \sqrt{16} \times \sqrt{3}} \\ & = {3\sqrt{2} - 2\sqrt{3} \over 5\sqrt{2} - 4\sqrt{3}} \\ & = {3\sqrt{2} - 2\sqrt{3} \over 5\sqrt{2} - 4\sqrt{3}} \times {5\sqrt{2} + 4\sqrt{3} \over 5\sqrt{2} + 4\sqrt{3}} \phantom{00000} [\text{Rationalise denominator}] \\ & = {(3\sqrt{2} - 2\sqrt{3})(5\sqrt{2} + 4\sqrt{3}) \over (5\sqrt{2} - 4\sqrt{3})(5\sqrt{2} + 4\sqrt{3})} \\ & = {(3\sqrt{2})(5\sqrt{2}) + (3\sqrt{2})(4\sqrt{3}) - (2\sqrt{3})(5\sqrt{2}) - (2\sqrt{3})(4\sqrt{3}) \over (5\sqrt{2})^2 - (4\sqrt{3})^2} \\ & = {30 + 12\sqrt{6} - 10\sqrt{6} - 24 \over 50 - 48} \\ & = {6 + 2\sqrt{6} \over 2} \\ & = {6 \over 2} + {2\sqrt{6} \over 2} \\ & = 3 + \sqrt{6} \end{align}

(a)

\begin{align} x\sqrt{3} - x & = 2 \\ x(\sqrt{3} - 1) & = 2 \\ x & = {2 \over \sqrt{3} - 1} \\ & = {2 \over \sqrt{3} - 1} \times {\sqrt{3} + 1 \over \sqrt{3} + 1} \phantom{00000} [\text{Rationalise denominator}] \\ & = { 2(\sqrt{3} + 1) \over (\sqrt{3} - 1)(\sqrt{3} + 1) } \\ & = {2\sqrt{3} + 2 \over (\sqrt{3})^2 - (1)^2} \\ & = {2\sqrt{3} + 2 \over 3 - 1} \\ & = {2\sqrt{3} + 2 \over 2} \\ & = {2\sqrt{3} \over 2} + {2 \over 2} \\ & = \sqrt{3} + 1 \end{align}

(b)

\begin{align} \sqrt{2x - 3} & = 5 \\ (\sqrt{2x - 3})^2 & = (5)^2 \\ 2x - 3 & = 25 \\ 2x & = 28 \\ x & = 14 \\ \\ \text{For } & x = 14, \\ \text{L.H.S} & = \sqrt{2(14) -3} = 5 = \text{R.H.S} \\ \\ \therefore x & = 14 \end{align}

(c)

\begin{align} 2\sqrt{1 - x} & = \sqrt{3} \\ (2\sqrt{1-x})^2 & = (\sqrt{3})^2 \\ (4)(1 - x) & = 3 \\ 1 - x & = {3 \over 4} \\ 1 - x & = 0.75 \\ 1 - 0.75 & = x \\ {1 \over 4} & = x \\ \\ \text{For } & x = {1 \over 4}, \\ \text{L.H.S} & = 2\sqrt{1 - {1 \over 4}} \\ & = 2 \sqrt{3 \over 4} \\ & = 2 { \sqrt{3} \over \sqrt{4} } \\ & = 2 { \sqrt{3} \over 2} \\ & = \sqrt{3} = \text{R.H.S} \\ \\ \therefore x & = {1 \over 4} \end{align}

(d)

\begin{align} \sqrt{8 - x} & = \sqrt{x - 2} \\ (\sqrt{8 - x})^2 & = (\sqrt{x - 2})^2 \\ 8 - x & = x - 2 \\ - 2x & = -10 \\ x & = 5 \\ \\ \text{For } & x = 5, \\ \text{L.H.S} & = \sqrt{8 - 5} = \sqrt{3} \\ \text{R.H.S} & = \sqrt{5 - 2} = \sqrt{3} \\ \\ \therefore x & = 5 \end{align}

(a)

\begin{align} a + b\sqrt{3} & = 2 - \sqrt{3} \\ & = 2 + (-1)\sqrt{3} \\ \\ \therefore & \phantom{0} a = 2, b = -1 \end{align}

(b)

\begin{align} a + b\sqrt{7} & = 4\sqrt{7} - 12 - \sqrt{7} \\ & = 4\sqrt{7} - \sqrt{7} - 12 \\ & = \sqrt{7}(4 - 1) - 12 \\ & = 3\sqrt{7} - 12 \\ \\ a + b\sqrt{7} & = -12 + 3\sqrt{7} \\ \\ \therefore & \phantom{0} a = -12, b = 3 \end{align}

(c)

\begin{align} a + b\sqrt{2} & = 3(1 - \sqrt{2}) + 4\sqrt{2} \\ & = 3 - 3\sqrt{2} + 4\sqrt{2} \\ & = 3 - \sqrt{2} (3 - 4) \\ & = 3 - \sqrt{2} (- 1) \\ & = 3 + \sqrt{2} \\ \\ \therefore & \phantom{0} a = 3, b = 1 \end{align}

(i)

\begin{align} \text{When } & s = 40, \\ v & = \sqrt{20(40)} \\ & = \sqrt{800} \\ & = \sqrt{400} \times \sqrt{2} \\ & = 20\sqrt{2} \end{align}

(ii)

\begin{align} \text{When } & s = 90, \\ v & = \sqrt{20(90)} \\ & = \sqrt{1800} \\ & = \sqrt{900} \times \sqrt{2} \\ & = 30\sqrt{2} \end{align}

(iii)

\begin{align} \text{Change in velocity } & = 30\sqrt{2} - 20\sqrt{2} \\ & = \sqrt{2} (30 - 20) \\ & = 10\sqrt{2} \text{ m/s} \end{align}

(i)

\begin{align} \text{Total length } & = 8 \times (3 + 2\sqrt{5}) + 4 \times (\sqrt{80} - 3) \\ & = 8(3 + 2\sqrt{5}) + 4(\sqrt{80} - 3) \\ & = 24 + 16\sqrt{5} + 4(\sqrt{16 \times 5} -3) \\ & = 24 + 16\sqrt{5} + 4(\sqrt{16} \times \sqrt{5} - 3) \\ & = 24 + 16\sqrt{5} + 4(4\sqrt{5} - 3) \\ & = 24 + 16\sqrt{5} + 16\sqrt{5} - 12 \\ & = (12 + 32\sqrt{5}) \text{ cm} \end{align}

(ii)

\begin{align} \text{Volume } & = \text{Base area } \times \text{Height} \\ & = (3 + 2\sqrt{5})(3 + 2\sqrt{5}) \times (\sqrt{80} - 3) \\ & = [(3)^2 + 2(3)(2\sqrt{5}) + (2\sqrt{5})^2] \times (\sqrt{16} \times \sqrt{5} - 3) \\ & = (9 + 12\sqrt{5} + 20) \times (4\sqrt{5} - 3) \\ & = (29 + 12\sqrt{5})(4\sqrt{5} - 3) \\ & = 116\sqrt{5} - 87 + (12\sqrt{5})(4\sqrt{5}) - 36\sqrt{5} \\ & = 80\sqrt{5} - 87 + 240 \\ & = (80\sqrt{5} + 153) \text{ cm}^3 \end{align}

(a)

\begin{align} 4\sqrt{12} - {2 \over 3}\sqrt{3} - {18 \over \sqrt{3}} & = 4\sqrt{4 \times 3} - {2 \over 3}\sqrt{3} - {18 \over \sqrt{3}}\times {\sqrt{3} \over \sqrt{3}} \\ & = 4(\sqrt{4})(\sqrt{3}) - {2 \over 3}\sqrt{3} - {18\sqrt{3} \over 3} \\ & = 8\sqrt{3} - {2 \over 3}\sqrt{3} - 6\sqrt{3} \\ & = {4 \over 3}\sqrt{3} \end{align}

(b)

\begin{align} {4 \over \sqrt{2}} - {\sqrt{128} \over 3} + {2 \over \sqrt{8}} & = {4 \over \sqrt{2}} \times {\sqrt{2} \over \sqrt{2}} - {\sqrt{64 \times 2} \over 3} + {2 \over \sqrt{8}}\times {\sqrt{8} \over \sqrt{8}} \\ & = {4\sqrt{2} \over 2} - {\sqrt{64}(\sqrt{2}) \over 3} + {2\sqrt{8} \over 8} \\ & = 2\sqrt{2} - {8\sqrt{2} \over 3} + {\sqrt{8} \over 4} \\ & = 2\sqrt{2} - {8 \over 3}\sqrt{2} + {\sqrt{4 \times 2} \over 4} \\ & = -{2 \over 3}\sqrt{2} + {\sqrt{4}(\sqrt{2}) \over 4} \\ & = -{2 \over 3}\sqrt{2} + {2\sqrt{2} \over 4} \\ & = -{2 \over 3}\sqrt{2} + {\sqrt{2} \over 2} \\ & = -{2 \over 3}\sqrt{2} + {1 \over 2}\sqrt{2} \\ & = -{1 \over 6}\sqrt{2} \end{align}

(c)

\begin{align} {2 \over \sqrt{6}} & = {2 \over \sqrt{6}} \times {\sqrt{6} \over \sqrt{6}} \\ & = {2\sqrt{6} \over 6} \\ & = {\sqrt{6} \over 3} \\ & = {1 \over 3} \sqrt{6} \\ & = {1 \over 3} \sqrt{2 \times 3} \\ & = {1 \over 3}\sqrt{2}\sqrt{3} \\ \\ {4 \over \sqrt{27}} & = {4 \over \sqrt{27}} \times {\sqrt{27} \over \sqrt{27}} \\ & = {4\sqrt{27} \over 27} \\ & = {4 \over 27} \sqrt{27} \\ & = {4 \over 27} \sqrt{9 \times 3} \\ & = {4 \over 27}(\sqrt{9})(\sqrt{3}) \\ & = {4 \over 27}(3)(\sqrt{3}) \\ & = {4 \over 9} \sqrt{3} \\ \\ { \sqrt{243} \over 2} & = {\sqrt{81 \times 3} \over 2} \\ & = {(\sqrt{81})(\sqrt{3}) \over 2} \\ & = {9\sqrt{3} \over 2} \\ & = {9 \over 2}\sqrt{3} \\ \\ {5 \over \sqrt{12}} & = {5 \over \sqrt{12}} \times {\sqrt{12} \over \sqrt{12}} \\ & = {5\sqrt{12} \over 12} \\ & = {5 \over 12}\sqrt{12} \\ & = {5 \over 12}\sqrt{4 \times 3} \\ & = {5 \over 12}(\sqrt{4})(\sqrt{3}) \\ & = {5 \over 12}(2)(\sqrt{3}) \\ & = {5 \over 6}\sqrt{3} \\ \\ \\ \therefore {2 \over \sqrt{6}}\left( {4 \over \sqrt{27}} + {\sqrt{243} \over 2} - {5 \over \sqrt{12}} \right) & = {1 \over 3}\sqrt{2}\sqrt{3} \left( {4 \over 9} \sqrt{3} + {9 \over 2}\sqrt{3} - {5 \over 6}\sqrt{3} \right) \\ & = {1 \over 3}\sqrt{2}\sqrt{3} \left( {37 \over 9} \sqrt{3} \right) \\ & = {1 \over 3}\left(37 \over 9\right)(\sqrt{2})(\sqrt{3})(\sqrt{3}) \\ & = {1 \over 3}\left(37 \over 9\right)(\sqrt{2})(3) \\ & = \left(37 \over 9\right)(\sqrt{2}) \\ & = {37 \over 9}\sqrt{2} \end{align}

(a)

\begin{align} {1 \over 4\sqrt{3} - 2} & = {1 \over 4\sqrt{3} - 2} \times {4\sqrt{3} + 2 \over 4\sqrt{3} + 2} \\ & = {4\sqrt{3} + 2 \over (4\sqrt{3} - 2)(4\sqrt{3} + 2)} \\ & = {4\sqrt{3} + 2 \over (4\sqrt{3})^2 - (2)^2} \\ & = {4\sqrt{3} + 2 \over 48 - 4} \\ & = {4\sqrt{3} + 2 \over 44} \\ \\ {2 \over 4\sqrt{3} + 2} & = {2 \over 4\sqrt{3} + 2} \times {4\sqrt{3} - 2 \over 4\sqrt{3} - 2} \\ & = {2(4\sqrt{3} - 2) \over (4\sqrt{3} + 2)(4\sqrt{3} - 2)} \\ & = {8\sqrt{3} - 4 \over (4\sqrt{3})^2 - (2)^2} \\ & = {8\sqrt{3} - 4 \over 48 - 4} \\ & = {8\sqrt{3} - 4 \over 44} \\ \\ \\ \therefore {1 \over 4\sqrt{3} - 2} - {2 \over 4\sqrt{3} + 2} & = {4\sqrt{3} + 2 \over 44} - {8\sqrt{3} - 4 \over 44} \\ & = {4\sqrt{3} + 2 - (8\sqrt{3} - 4) \over 44} \\ & = {4\sqrt{3} + 2 - 8\sqrt{3} + 4 \over 44} \\ & = {6 - 4\sqrt{3} \over 44} \\ & = {6 \over 44} - {4\sqrt{3} \over 44} \\ & = {3 \over 22} - {\sqrt{3} \over 11} \\ & = {3 \over 22} - {1 \over 11}\sqrt{3} \end{align}

(b)

\begin{align} {1 \over (2\sqrt{7} - 3)^2} & = {1 \over (2\sqrt{7})^2 - 2(2\sqrt{7})(3) + (3)^2} \\ & = {1 \over 28 - 12\sqrt{7} + 9} \\ & = {1 \over 37 - 12\sqrt{7}} \times {37 + 12\sqrt{7} \over 37 + 12\sqrt{7}} \\ & = {37 + 12\sqrt{7} \over (37)^2 - (12\sqrt{7})^2} \\ & = {37 + 12\sqrt{7} \over 1369 - 1008} \\ & = {37 + 12\sqrt{7} \over 361} \\ \\ {1 \over (2\sqrt{7} + 3)^2} & = {1 \over (2\sqrt{7})^2 + 2(2\sqrt{7})(3) + (3)^2} \\ & = {1 \over 28 + 12\sqrt{7} + 9} \\ & = {1 \over 37 + 12\sqrt{7}} \times {37 - 12\sqrt{7} \over 37 - 12\sqrt{7}} \\ & = {37 - 12\sqrt{7} \over (37)^2 - (12\sqrt{7})^2} \\ & = {37 - 12\sqrt{7} \over 1369 - 1008} \\ & = {37 - 12\sqrt{7} \over 361} \\ \\ \\ \therefore {1 \over (2\sqrt{7} - 3)^2} - {1 \over (2\sqrt{7} + 3)^2} & = {37 + 12\sqrt{7} \over 361} - {37 - 12\sqrt{7} \over 361} \\ & = {37 + 12\sqrt{7} - (37 - 12\sqrt{7}) \over 361} \\ & = {37 + 12\sqrt{7} - 37 + 12\sqrt{7} \over 361} \\ & = {24\sqrt{7} \over 361} \\ & = {24 \over 361}\sqrt{7} \end{align}

(c)

\begin{align} \sqrt{6}(4\sqrt{2} + 3\sqrt{3}) & = (\sqrt{3 \times 2})(4\sqrt{2} + 3\sqrt{3}) \\ & = (\sqrt{3})(\sqrt{2})(4\sqrt{2} + 3\sqrt{3}) \\ & = (\sqrt{3})(8 + 3\sqrt{2}\sqrt{3}) \\ & = 8\sqrt{3} + 9\sqrt{2} \\ \\ {38 \over 5\sqrt{2} - 2\sqrt{3}} & = {38 \over 5\sqrt{2} - 2\sqrt{3}} \times {5\sqrt{2} + 2\sqrt{3} \over 5\sqrt{2} + 2\sqrt{3}} \\ & = {38(5\sqrt{2} + 2\sqrt{3}) \over (5\sqrt{2} - 2\sqrt{3})(5\sqrt{2} + 2\sqrt{3})} \\ & = {190\sqrt{2} + 76\sqrt{3} \over (5\sqrt{2})^2 - (2\sqrt{3})^2} \\ & = {190\sqrt{2} + 76\sqrt{3} \over 50 - 12} \\ & = {190\sqrt{2} + 76\sqrt{3} \over 38} \\ & = {190\sqrt{2} \over 38} + {76\sqrt{3} \over 38} \\ & = 5\sqrt{2} + 2\sqrt{3} \\ \\ \\ \therefore \sqrt{6} (4\sqrt{2} + 3\sqrt{3}) - {38 \over 5\sqrt{2} - 2\sqrt{3}} & = 8\sqrt{3} + 9\sqrt{2} - (5\sqrt{2} + 2\sqrt{3}) \\ & = 8\sqrt{3} + 9\sqrt{2} - 5\sqrt{2} - 2\sqrt{3} \\ & = 6\sqrt{3} + 4\sqrt{2} \end{align}

(a)

\begin{align} (3 - \sqrt{3})^2 - {5 \over 2- \sqrt{3}} & = [(3)^2 - 2(3)(\sqrt{3}) + (\sqrt{3})^2] - {5 \over 2 - \sqrt{3}} \\ & = 9 - 6\sqrt{3} + 3 - {5 \over 2 - \sqrt{3}} \\ & = 12 - 6\sqrt{3} - {5 \over 2 - \sqrt{3}} \times {2 + \sqrt{3} \over 2 + \sqrt{3}} \phantom{00000} [\text{Rationalise denominator]} \\ & = 12 - 6\sqrt{3} - {5(2 + \sqrt{3}) \over (2 - \sqrt{3})(2 + \sqrt{3})} \\ & = 12 - 6\sqrt{3} - {10 + 5\sqrt{3} \over (2)^2 - (\sqrt{3})^2} \\ & = 12 - 6\sqrt{3} - {10 + 5\sqrt{3} \over 4 - 3} \\ & = 12 - 6\sqrt{3} - {10 + 5\sqrt{3} \over 1} \\ & = 12 - 6\sqrt{3} - (10 + 5\sqrt{3}) \\ & = 12 - 6\sqrt{3} - 10 - 5\sqrt{3} \\ & = 2 - 11\sqrt{3} \end{align}

(b)

\begin{align} {s^2 + 1 \over s + 2} & = {(1 - \sqrt{5})^2 + 1 \over (1 - \sqrt{5}) + 2} \\ & = {(1)^2 - 2(1)(\sqrt{5}) + (\sqrt{5})^2 + 1 \over 1 - \sqrt{5} + 2} \\ & = {1 - 2\sqrt{5} + 5 + 1 \over 3 - \sqrt{5}} \\ & = {7 - 2\sqrt{5} \over 3 - \sqrt{5}} \\ & = {7 - 2\sqrt{5} \over 3 - \sqrt{5}} \times {3 + \sqrt{5} \over 3 + \sqrt{5}} \phantom{00000} [\text{Rationalise denominator]} \\ & = {(7 - 2\sqrt{5})(3 + \sqrt{5}) \over (3 - \sqrt{5})(3 + \sqrt{5})} \\ & = {21 + 7\sqrt{5} - 6\sqrt{5} - 10 \over (3)^2 - (\sqrt{5})^2} \\ & = {11 + \sqrt{5} \over 9 - 5} \\ & = {11 + \sqrt{5} \over 4} \\ & = {11 \over 4} + {\sqrt{5} \over 4} \\ & = {11 \over 4} + {1 \over 4}\sqrt{5} \end{align}

(i)

\begin{align} \text{Length } & = 5 - \sqrt{12} \\ & = 5 - (\sqrt{4})(\sqrt{3}) \\ & = 5 - 2\sqrt{3} \\ \\ \text{Breadth } & = 4 + {6 \over \sqrt{3}} \\ & = 4 + {6 \over \sqrt{3}} \times {\sqrt{3} \over \sqrt{3}} \\ & = 4 + {6\sqrt{3} \over 3} \\ & = 4 + 2\sqrt{3} \end{align}
\begin{align} \text{Area } & = \text{Length } \times \text{Breadth} \\ & = (5 - 2\sqrt{3})(4 + 2\sqrt{3}) \\ & = 20 + 10\sqrt{3} - 8\sqrt{3} - (2\sqrt{3})^2 \\ & = 20 + 2\sqrt{3} - 12 \\ & = (8 + 2\sqrt{3}) \text{ m}^2 \end{align}

(ii)

\begin{align} \text{By Pythago} & \text{ras theorem,} \\ \text{Diagonal }^2 & = \text{Length}^2 + \text{Breadth}^2 \\ & = (5 - 2\sqrt{3})^2 + (4 + 2\sqrt{3})^2 \\ & = [(5)^2 - 2(5)(2\sqrt{3}) + (2\sqrt{3})^2] + [(4)^2 + 2(4)(2\sqrt{3}) + (2\sqrt{3})^2] \\ & = (25 - 20\sqrt{3} + 12) + (16 + 16\sqrt{3} + 12) \\ & = (37 - 20\sqrt{3}) + (28 + 16\sqrt{3}) \\ & = 37 - 20\sqrt{3} + 28 + 16\sqrt{3} \\ & = 65 - 4\sqrt{3} \end{align}

(a)

\begin{align} \sqrt{2 + x} - x & = 0 \\ \sqrt{2 + x} & = x \\ \\ (\sqrt{2 + x})^2 & = (x)^2 \\ 2 + x & = x^2 \\ 0 & = x^2 - x - 2 \\ 0 & = (x - 2)(x + 1) \end{align} \begin{align} x - 2 & = 0 & x + 1 & = 0 \\ x & = 2 & x & = - 1 \end{align} \begin{align} \text{For } & x = 2, \\ \text{L.H.S } & = \sqrt{2 + (2)} - (2) \\ & = 0 = \text{R.H.S} \\ \\ \text{For } & x = -1, \\ \text{L.H.S } & = \sqrt{2 + (-1)} - (-1) \\ & = 2 \ne \text{R.H.S} \\ \\ \text{Reject } x & = -1 \\ \\ \therefore x & = 2 \end{align}

(b)

\begin{align} \sqrt{6 - 5x} - x & = - 2x \\ \sqrt{6 - 5x} & = - x \\ \\ (\sqrt{6 - 5x})^2 & = (-x)^2 \\ 6 - 5x & = x^2 \\ 0 & = x^2 + 5x - 6 \\ 0 & = (x + 6)(x - 1) \end{align} \begin{align} x + 6 & = 0 & x - 1 & = 0 \\ x & = -6 & x & = 1 \end{align} \begin{align} \text{For } & x = -6, \\ \text{L.H.S } & = \sqrt{6 - 5(-6)} - (-6) = 12 \\ \text{R.H.S} & = -2(-6) = 12 \\ \\ \text{For } & x = 1, \\ \text{L.H.S} & = \sqrt{6 - 5(1)} - (1) = 0 \\ \text{R.H.S} & = -2(1) = -2 \\ \\ \text{Reject } x & = 1 \\ \\ \therefore x & = -6 \end{align}

(c)

\begin{align} \sqrt{x + 5} + x & = 1 \\ \sqrt{x + 5} & = 1 - x \\ \\ (\sqrt{x + 5})^2 & = (1 - x)^2 \\ x + 5 & = 1 - 2x + x^2 \\ 0 & = x^2 - 3x - 4 \\ 0 & = (x - 4)(x + 1) \end{align} \begin{align} x - 4 & = 0 & x + 1 & = 0 \\ x & = 4 & x & = -1 \end{align} \begin{align} \text{For } & x = 4, \\ \text{L.H.S } & = \sqrt{(4) + 5} + (4) - 1 \\ & = 6 \ne \text{R.H.S} \\ \\ \text{Reject } x & = 4 \\ \\ \text{For } & x = -1, \\ \text{L.H.S } & = \sqrt{(-1) + 5} + (-1) - 1 \\ & = 0 = \text{R.H.S} \\ \\ \therefore x & = -1 \end{align}

\begin{align} x\sqrt{40} & = x\sqrt{5} + \sqrt{10} \\ x\sqrt{40} - x\sqrt{5} & = \sqrt{10} \\ x(\sqrt{40} - \sqrt{5}) & = \sqrt{10} \\ x & = {\sqrt{10} \over \sqrt{40} - \sqrt{5}} \\ & = {\sqrt{10} \over \sqrt{40} - \sqrt{5}} \times {\sqrt{40} + \sqrt{5} \over \sqrt{40} + \sqrt{5}} \phantom{00000} [\text{Rationalise denominator]} \\ & = { \sqrt{10}(\sqrt{40} + \sqrt{5}) \over (\sqrt{40} - \sqrt{5})(\sqrt{40} + \sqrt{5}) } \\ & = { \sqrt{400} + \sqrt{50} \over (\sqrt{40})^2 - (\sqrt{5})^2} \\ & = { 20 + \sqrt{50} \over 40 - 55 } \\ & = { 20 + \sqrt{50} \over 35} \\ & = { 20 + \sqrt{25 \times 2} \over 35 } \\ & = { 20 + \sqrt{25}(\sqrt{2}) \over 35 } \\ & = { 20 + 5\sqrt{2} \over 35} \\ & = { 5(4 + \sqrt{2}) \over 35} \\ & = {4 + \sqrt{2} \over 7} \\ \\ \therefore a = 4, & \phantom{.} b = 2 \end{align}

(a)

\begin{align} 3 + a\sqrt{5} & = b(3 - \sqrt{5}) - 2(3 + \sqrt{5}) \\ & = 3b - b\sqrt{5} - 6 - 2\sqrt{5} \\ & = 3b - 6 - b\sqrt{5} - 2\sqrt{5} \\ & = 3b - 6 - (b\sqrt{5} + 2\sqrt{5}) \\ \\ 3 + a\sqrt{5} & = (3b - 6) - (b +2)\sqrt{5} \\ \\ \text{Comparing} & \text{ coefficients,} \\ 3 & = 3b - 6 \\ 9 & = 3b \\ {9 \over 3} & = b \\ 3 & = b \\ \\ a & = -(b + 2) \\ a & = - b - 2 \\ a & = -(3) - 2 \\ a & = -5 \\ \\ \therefore a & = -5, b = 3 \end{align}

(b)

\begin{align} a - b\sqrt{7} & = (2 + \sqrt{7})(2 - \sqrt{7}) + \sqrt{7} \\ & = (2)^2 - (\sqrt{7})^2 + \sqrt{7} \\ & = 4 - 7 + \sqrt{7} \\ & = -3 + \sqrt{7} \\ & = (-3) - (-1)\sqrt{7} \\ \\ \therefore a & = -3, b = -1 \end{align}

(c)

\begin{align} a + \sqrt{2} & = (3 - 2\sqrt{2})^2 + b\sqrt{8} \\ & = (3)^2 - 2(3)(2\sqrt{2}) + (2\sqrt{2})^2 + b\sqrt{8} \\ & = 9 - 12\sqrt{2} + 8 + b\sqrt{8} \\ & = 17 - 12\sqrt{2} + b\sqrt{8} \\ & = 17 - 12\sqrt{2} + b\sqrt{4 \times 2} \\ & = 17 - 12\sqrt{2} + b(\sqrt{4})(\sqrt{2}) \\ & = 17 - 12\sqrt{2} + 2b\sqrt{2} \\ & = 17 + 2b\sqrt{2} - 12\sqrt{2} \\ \\ a + \sqrt{2} & = 17 + (2b - 12)\sqrt{2} \\ \\ \text{Comparing} & \text{ coefficients,} \\ a & = 17 \\ \\ 1 & = 2b - 12 \\ 13 & = 2b \\ {13 \over 2} & = b \\ \\ \therefore a & = 17, b = {13 \over 2} \end{align}

\begin{align} {18 \over 4 -\sqrt{7}} & = {18 \over 4 - \sqrt{7}} \times {4 + \sqrt{7} \over 4 + \sqrt{7}} \\ & = {18(4 + \sqrt{7}) \over (4 - \sqrt{7})(4 + \sqrt{7}) } \\ & = {72 + 18\sqrt{7} \over (4)^2 - (\sqrt{7})^2} \\ & = {72 + 18\sqrt{7} \over 16 - 7} \\ & = {72 + 18\sqrt{7} \over 9} \\ & = 8 + 2\sqrt{7} \\ \\ \therefore \sqrt{a + b\sqrt{7}} & = 8 + 2\sqrt{7} \\ (\sqrt{a + b\sqrt{7}})^2 & = (8 + 2\sqrt{7})^2 \\ a + b\sqrt{7} & = (8)^2 + 2(8)(2\sqrt{7}) + (2\sqrt{7})^2 \\ & = 64 + 32\sqrt{7} + 28 \\ & = 92 + 32\sqrt{7} \\ \\ \therefore a & = 92, b = 32 \end{align}

(a)

\begin{align} (a + 3\sqrt{2})(3 - 4\sqrt{2}) & = -18 + b\sqrt{2} \\ 3a - 4a\sqrt{2} + 9\sqrt{2} -24 & = -18 + b\sqrt{2} \\ 3a - 24 + 9\sqrt{2} - 4a\sqrt{2} & = -18 + b\sqrt{2} \\ \\ (3a - 24) + (9 - 4a)\sqrt{2} & = -18 + b\sqrt{2} \\ \\ \text{Comparing} & \text{ coefficients,} \\ 3a - 24 & = -18 \\ 3a & = 6 \\ a & = {6 \over 3} \\ a & = 2 \\ \\ 9 - 4a & = b \\ 9 - 4(2) & = b \\ 1 & = b \\ \\ \therefore a & = 2, b = 1 \end{align}

(b)

\begin{align} (3 - 2\sqrt{3})(4 + a\sqrt{3}) & = b - 23\sqrt{3} \\ 12 + 3a\sqrt{3} - 8\sqrt{3} - 6a & = b - 23\sqrt{3} \\ 12 - 6a + 3a\sqrt{3} - 8\sqrt{3} & = b - 23\sqrt{3} \\ \\ (12 - 6a) + (3a - 8)\sqrt{3} & = b - 23\sqrt{3} \\ \\ \text{Comparing} & \text{ coefficients,} \\ 3a - 8 & = - 23 \\ 3a & = -15 \\ a & = {-15 \over 3} \\ a & = -5 \\ \\ 12 - 6a & = b \\ 12 - 6(-5) & = b \\ 42 & = b \\ \\ \therefore a & = -5, b = 42 \end{align}

(c)

\begin{align} (a + b\sqrt{3}) (5 - 3\sqrt{3}) & = 6 - 4\sqrt{3} \\ \\ a + b \sqrt{3} & = {6 - 4\sqrt{3} \over 5 - 3\sqrt{3}} \\ & = {6 - 4\sqrt{3} \over 5 - 3\sqrt{3}} \times {5 + 3\sqrt{3} \over 5 + 3\sqrt{3}} \\ & = {(6 - 4\sqrt{3})(5 + 3\sqrt{3}) \over (5 - 3\sqrt{3})(5 + 3\sqrt{3})} \\ & = {30 + 18\sqrt{3} - 20\sqrt{3} - 36 \over (5)^2 - (3\sqrt{3})^2} \\ & = {-6 - 2\sqrt{3} \over 25 - 27} \\ & = {-6 - 2\sqrt{3} \over -2} \\ & = {-6 \over -2} - {2\sqrt{3} \over -2} \\ & = 3 - (-\sqrt{3}) \\ & = 3 + \sqrt{3} \\ \\ a + b\sqrt{3}& = 3 + (1)\sqrt{3} \\ \\ \therefore a = 3 &, b = 1 \end{align}

(d)

\begin{align} {2 + 4\sqrt{7} \over a + b\sqrt{7}} & = 4 - \sqrt{7} \\ 2 + 4\sqrt{7} & = (4 - \sqrt{7})(a + b\sqrt{7}) \\ {2 + 4\sqrt{7} \over 4 - \sqrt{7}} & = a + b\sqrt{7} \\ \\ a + b\sqrt{7} & = {2 + 4\sqrt{7} \over 4 - \sqrt{7}} \times {4 + \sqrt{7} \over 4 + \sqrt{7}} \\ & = { (2 + 4\sqrt{7})(4 + \sqrt{7}) \over (4 - \sqrt{7})(4 + \sqrt{7}) } \\ & = { 8 + 2\sqrt{7} + 16\sqrt{7} + 28 \over (4)^2 - (\sqrt{7})^2} \\ & = { 36 + 18\sqrt{7} \over 16 - 7 } \\ & = { 36 + 18\sqrt{7} \over 9} \\ & = {36 \over 9} + {18\sqrt{7} \over 9} \\ & = 4 + 2\sqrt{7} \\ \\ a + b\sqrt{7} & = 4 + 2\sqrt{7} \\ \\ \therefore a = 4 &, b = 2 \end{align}

(i)

\begin{align} (2 - \sqrt{3})^2 & = (2)^2 - 2(2)(\sqrt{3}) + (\sqrt{3})^2 \\ & = 4 - 4\sqrt{3} + 3 \\ & = 7 - 4\sqrt{3} \end{align}

(ii)

\begin{align} \pm\sqrt{28 - 16\sqrt{3}} & = \pm\sqrt{4(7 - 4\sqrt{3})} \\ & = \pm \sqrt{4} \underbrace{\sqrt{7-4\sqrt{3}}}_{\text{Part (i)}} \\ & = \pm\sqrt{4}\sqrt{(2 - \sqrt{3})^2} \\ & = \pm(2)(2 - \sqrt{3}) \\ & = \pm(4 - 2\sqrt{3}) \end{align}

(iii) Using parts (i) and (ii) as reference,

\begin{align} 34 - 24\sqrt{2} & = 2(17 - 12\sqrt{2}) \\ & = 2[17 - 2(3)(2)\sqrt{2}] \\ & = 2[9 + 8 - 2(3)(2\sqrt{2})] \\ & = 2[9 - 2(3)(2\sqrt{2}) + 8] \\ & = 2[(3)^2 - 2(3)(2\sqrt{2}) + (2\sqrt{2})^2] \\ & = 2(3 - 2\sqrt{2})^2 \\ \\ \therefore \sqrt{34 - 24\sqrt{2}} & = \pm\sqrt{2(3 - 2\sqrt{2})^2} \\ & = \pm \sqrt{2}\sqrt{(3 - 2\sqrt{2})^2} \\ & = \pm \sqrt{2}(3 - 2\sqrt{2}) \\ & = \pm (3\sqrt{2} - 4) \\ & = \pm (4 - 3\sqrt{2}) \end{align}

(i)

\begin{align} V & = IR \\ & = (5\sqrt{6}) \left[ {1 \over 10}(6\sqrt{2} + 7\sqrt{3}) \right] \\ & = {\sqrt{6} \over 2} (6\sqrt{2} + 7\sqrt{3}) \\ & = {\sqrt{3 \times 2} \over 2} (6\sqrt{2} + 7\sqrt{3}) \\ & = {\sqrt{3}\sqrt{2} \over 2} (6\sqrt{2} + 7\sqrt{3}) \\ & = {12\sqrt{3} \over 2} + {21\sqrt{2} \over 2} \\ & = 6\sqrt{3} + {21 \over 2}\sqrt{2} \end{align}

(ii)

\begin{align} {1 \over R} & = {1 \over {1 \over 10}(6\sqrt{2} + 7\sqrt{3})} \\ & = {10 \over 6\sqrt{2} + 7\sqrt{3}} \\ & = {10 \over 6\sqrt{2} + 7\sqrt{3}} \times {6\sqrt{2} - 7\sqrt{3} \over 6\sqrt{2} - 7\sqrt{3}} \\ & = {10(6\sqrt{2} - 7\sqrt{3}) \over (6\sqrt{2} + 7\sqrt{3})(6\sqrt{2} - 7\sqrt{3})} \\ & = {60\sqrt{2} - 70\sqrt{3} \over (6\sqrt{2})^2 - (7\sqrt{3})^2} \\ & = {60\sqrt{2} - 70\sqrt{3} \over 72 - 147} \\ & = {60\sqrt{2} - 70\sqrt{3} \over -75} \\ & = {12\sqrt{2} - 14\sqrt{3} \over -15} \\ & = {-(12\sqrt{2} - 14\sqrt{3}) \over 15} \\ & = {14\sqrt{3} - 12\sqrt{2} \over 15} \\ \\ {1 \over R_1} & = {1 \over \sqrt{3} + 3\sqrt{2}} \\ & = {1 \over \sqrt{3} + 3\sqrt{2}} \times {\sqrt{3} - 3\sqrt{2} \over \sqrt{3} - 3\sqrt{2}} \\ & = {\sqrt{3} - 3\sqrt{2} \over (\sqrt{3} + 3\sqrt{2})(\sqrt{3} - 3\sqrt{2})} \\ & = {\sqrt{3} - 3\sqrt{2} \over (\sqrt{3})^2 - (3\sqrt{2})^2} \\ & = {\sqrt{3} - 3\sqrt{2} \over 3 - 18} \\ & = {\sqrt{3} - 3\sqrt{2} \over -15} \\ & = -{\sqrt{3} - 3\sqrt{2} \over 15} \\ \\ \\ {1 \over R} & = {1 \over R_1} + {1 \over R_2} \\ \\ {1 \over R_2} & = {1 \over R} - {1 \over R_1} \\ & = {14\sqrt{3} - 12\sqrt{2} \over 15} - \left(- {\sqrt{3} - 3\sqrt{2} \over 15} \right) \\ & = {14\sqrt{3} - 12\sqrt{2} \over 15} + {\sqrt{3} - 3\sqrt{2} \over 15} \\ & = {14\sqrt{3} - 12\sqrt{2} + \sqrt{3} - 3\sqrt{2} \over 15} \\ & = {15\sqrt{3} - 15\sqrt{2} \over 15} \\ & = {15\sqrt{3} \over 15} - {15\sqrt{2} \over 15} \\ & = \sqrt{3} - \sqrt{2} \\ \\ {1 \over R_2} & = {\sqrt{3} - \sqrt{2} \over 1} \\ {1 \over \sqrt{3} - \sqrt{2}} & = R_2 \\ \\ R_2 & = {1 \over \sqrt{3} - \sqrt{2}} \times {\sqrt{3} + \sqrt{2} \over \sqrt{3} + \sqrt{2}} \\ & = {\sqrt{3} + \sqrt{2} \over (\sqrt{3})^2 - \sqrt{2})^2} \\ & = {\sqrt{3} + \sqrt{2} \over 1} \\ & = \sqrt{3} + \sqrt{2} \end{align}

(i)

\begin{align} {1 \over \sqrt{a} + \sqrt{a + 1}} & = {1 \over \sqrt{a} + \sqrt{a + 1}} \times {\sqrt{a} - \sqrt{a + 1} \over \sqrt{a} - \sqrt{a + 1}} \\ & = {\sqrt{a} - \sqrt{a + 1} \over (\sqrt{a} + \sqrt{a + 1})(\sqrt{a} - \sqrt{a + 1})} \\ & = {\sqrt{a} - \sqrt{a + 1} \over (\sqrt{a})^2 - (\sqrt{a + 1})^2} \\ & = {\sqrt{a} - \sqrt{a + 1} \over a - (a + 1) } \\ & = {\sqrt{a} - \sqrt{a + 1} \over a - a - 1 } \\ & = {\sqrt{a} - \sqrt{a + 1} \over - 1 } \\ & = {\sqrt{a} \over -1} - {\sqrt{a + 1} \over -1} \\ & = -\sqrt{a} - (-\sqrt{a + 1}) \\ & = \sqrt{a + 1} - \sqrt{a} \end{align}

(ii)

\begin{align} \text{First term, } {1 \over \sqrt{1} + \sqrt{2}} & = {1 \over \sqrt{1} + \sqrt{1 + 1} } \\ & = \sqrt{1 + 1} - \sqrt{1} \\ & = \sqrt{2} - \sqrt{1} \\ \\ \text{Second term, } {1 \over \sqrt{2} + \sqrt{3}} & = {1 \over \sqrt{2} + \sqrt{2 + 1} } \\ & = \sqrt{2 + 1} - \sqrt{2} \\ & = \sqrt{3} - \sqrt{2} \\ \\ \text{Third term, } {1 \over \sqrt{3} + \sqrt{4}} & = {1 \over \sqrt{3} + \sqrt{3 + 1} } \\ & = \sqrt{3 + 1} - \sqrt{3} \\ & = \sqrt{4} - \sqrt{3} \\ \end{align} \begin{align} & \therefore {1 \over \sqrt{1} + \sqrt{2}} + {1 \over \sqrt{2} + \sqrt{3}} + {1 \over \sqrt{3} + \sqrt{4}} + ... +{1 \over \sqrt{8} + \sqrt{9}} \\ = \phantom{0} & {1 \over \sqrt{8} + \sqrt{9}} + ... + {1 \over \sqrt{3} + \sqrt{4}} + {1 \over \sqrt{2} + \sqrt{3}} + {1 \over \sqrt{1} + \sqrt{2}} \\ = \phantom{0} & (\sqrt{9} - \sqrt{8}) + ... + (\sqrt{4} - \sqrt{3}) + (\sqrt{3} - \sqrt{2}) + (\sqrt{2} - \sqrt{1}) \\ = \phantom{0} & \sqrt{9} - \sqrt{1} \\ = \phantom{0} & 3 - 1 \\ = \phantom{0} & 2 \end{align}

\begin{align} \sqrt{x^2 - x - 2} + 3 \sqrt{x^2 + x - 6} & = 0 \\ \sqrt{(x - 2)(x + 1)} + 3 \sqrt{(x - 2)(x + 3)} & = 0 \\ \sqrt{x - 2} \sqrt{x + 1} + 3 \sqrt{x - 2} \sqrt{x + 3} & = 0 \\ \sqrt{x - 2} \left( \sqrt{x + 1} + 3 \sqrt{x + 3} \right) & = 0 \end{align}
\begin{align} \sqrt{x - 2} & = 0 && \text{ or } & \sqrt{x + 1} + 3 \sqrt{x + 3} & = 0 \\ x - 2 & = 0^2 &&& \sqrt{x + 1} & = - 3 \sqrt{x + 3} \phantom{0000} \text{ (Reject, since } \sqrt{x + 1} \ge 0) \\ x - 2 & = 0 \\ x & = 2 \end{align}