\begin{align} { \sqrt{20} + \sqrt{32} \over \sqrt{20} - \sqrt{32} } & = { \sqrt{4} \sqrt{5} + \sqrt{16} \sqrt{2} \over \sqrt{4} \sqrt{5} - \sqrt{16} \sqrt{2} } \\ & = {2 \sqrt{5} + 4 \sqrt{2} \over 2 \sqrt{5} - 4 \sqrt{2}} \times { 2 \sqrt{5} + 4 \sqrt{2} \over 2 \sqrt{5} + 4 \sqrt{2} } \\ & = { (2 \sqrt{5} + 4 \sqrt{2})^2 \over (2 \sqrt{5})^2 - (4 \sqrt{2})^2 } \phantom{0000000000000000000} [(a - b)(a + b) = a^2 - b^2] \\ & = { (2\sqrt{5})^2 + 2(2\sqrt{5})(4\sqrt{2}) + (4\sqrt{2})^2 \over -12} \phantom{000000} [(a + b)^2 = a^2 + 2ab + b^2] \\ & = { 20 + 16 \sqrt{10} + 32 \over -12} \\ & = { 52 + 16 \sqrt{10} \over -12} \\ & = { 52 \over -12} + {16 \sqrt{10} \over -12} \\ & = -{13 \over 3} - {4 \sqrt{10} \over 3} \\ & = -{13 \over 3} - {4 \over 3} \sqrt{10} \\ \\ \therefore a & = -{13 \over 3}, b = -{4 \over 3} \end{align}

\begin{align*} (a + \sqrt{3}) (10 - b\sqrt{27}) & = 10a - ab \sqrt{27} + 10\sqrt{3} - (\sqrt{3})(b\sqrt{27}) \\ & = 10a - ab (\sqrt{9} \times \sqrt{3}) + 10\sqrt{3} - b\sqrt{81} \\ & = 10a - ab (3)\sqrt{3} + 10\sqrt{3} - b(9) \\ & = 10a - 3ab \sqrt{3} + 10\sqrt{3} - 9b \\ & = 10a - 9b + (10 - 3ab)\sqrt{3} \\ \\ \text{Comparing with } & 28 + 16\sqrt{3}, \\ 10a - 9b & = 28 \phantom{000} \text{--- (1)} \\ \\ 10 - 3ab & = 16 \\ -3ab & = 16 - 10 \\ -3ab & = 6 \\ ab & = {6 \over -3} \\ ab & = -2 \\ a & = -{2 \over b} \phantom{000} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 10 \left(-{2 \over b}\right) - 9b & = 28 \\ -{20 \over b} - 9b & = 28 \\ -20- 9b^2 & = 28b \\ -9b^2 -28b - 20 & = 0 \\ 9b^2 + 28b + 20 & = 0 \\ (9b + 10)(b + 2) & = 0 \\ \\ 9b + 10 = 0 \phantom{000} & \text{ or } b + 2 = 0 \\ 9b = -10 \phantom{(} & \phantom{000000} b = - 2 \\ b = -{10 \over 9} & \phantom{000000} b = - 2 \\ \\ \text{Substitute } & b = -{10 \over 9} \text{ into (2),} \\ a & = -{2 \over -{10 \over 9}} \\ a & = {9 \over 5} \\ \\ \text{Substitute } & b = -2 \text{ into (2),} \\ a & = -{2 \over -2} \\ a & = 1 \end{align*}
$$ a = 1, b = -2 \text{ or } a = {9 \over 5}, b = - {10 \over 9} $$

\begin{align*} \text{Volume} & = (\sqrt{18} + \sqrt{48}) \pi \\ & = ( \sqrt{9} \times \sqrt{2} + \sqrt{16} \times \sqrt{3}) \pi \\ & = ( 3\sqrt{2} + 4\sqrt{3})\pi \\ \\ \text{Volume of cone} & = {1 \over 3} \pi r^2 h \\ (3\sqrt{2} + 4\sqrt{3})\pi & = {1 \over 3} \pi r^2 (2\sqrt{3} - \sqrt{2}) \\ (3\sqrt{2} + 4\sqrt{3}) & = {1 \over 3} r^2 (2\sqrt{3} - \sqrt{2}) \\ 3(3\sqrt{2} + 4\sqrt{3}) & = r^2 ( 2\sqrt{3} - \sqrt{2}) \\ 9\sqrt{2} + 12\sqrt{3} & = r^2 ( 2\sqrt{3} - \sqrt{2} ) \\ {9 \sqrt{2} + 12\sqrt{3} \over 2\sqrt{3} - \sqrt{2}} & = r^2 \end{align*} \begin{align*} r^2 & = { 9\sqrt{2} + 12\sqrt{3} \over 2\sqrt{3} - \sqrt{2}} \times {2\sqrt{3} + \sqrt{2} \over 2\sqrt{3} + \sqrt{2}} \\ & = { (9\sqrt{2} +12\sqrt{3})(2\sqrt{3} + \sqrt{2}) \over (2\sqrt{3} - \sqrt{2})(2\sqrt{3} + \sqrt{2})} \\ & = { (9\sqrt{2})(2\sqrt{3}) + (9\sqrt{2})(\sqrt{2}) + (12\sqrt{3})(2\sqrt{3}) + (12\sqrt{3})(\sqrt{2}) \over (2\sqrt{3})^2 - (\sqrt{2})^2 } \\ & = { 18\sqrt{6} + 9(2) + 24(3) + 12\sqrt{6} \over (4)(3) - 2} \\ & = { 30\sqrt{6} + 90 \over 10} \\ & = { 10( 3\sqrt{6} + 9 ) \over 10} \\ & = 3\sqrt{6} + 9 \end{align*} \begin{align*} \text{By Pythagoras theorem, } l^2 & = r^2 + h^2 \\ l^2 & = (3\sqrt{6} + 9) + (2\sqrt{3} - \sqrt{2})^2 \\ & = 3\sqrt{6} + 9 + (2\sqrt{3})^2 - 2(2\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 \\ & = 3\sqrt{6} + 9 + (4)(3) - 4\sqrt{6} + 2 \\ & = (23 - \sqrt{6}) \text{ cm}^2 \end{align*}

\begin{align} x^2 & + 6x - 10 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-6 \pm \sqrt{(6)^2 - 4(1)(-10)} \over 2(1)} \\ & = {-6 \pm \sqrt{76} \over 2} \\ & = {-6 \pm \sqrt{4} \sqrt{19} \over 2} \\ & = {-6 \pm 2 \sqrt{19} \over 2} \\ & = {-6 \over 2} \pm {2 \sqrt{19} \over 2} \\ & = -3 \pm \sqrt{19} \end{align}

\begin{align} x^2 + ax + b & = 0 \\ \\ \text{Let } x = -2 & + \sqrt{14}, \\ (-2 + \sqrt{14})^2 + a(-2 + \sqrt{14}) + b & = 0 \\ \underbrace{ (-2)^2 + 2(-2)(\sqrt{14}) + (\sqrt{14})^2}_{ (a + b)^2 = a^2 + 2ab + b^2} - 2a + a \sqrt{14} + b & = 0 \\ 4 - 4 \sqrt{14} + 14 & = 2a - b - a \sqrt{14} \\ 18 - 4 \sqrt{14} & = 2a - b - a \sqrt{14} \\ \\ \text{Comparing } & \text{coefficients,} \end{align} \begin{align} 18 & = 2a - b &&& -a & = -4 \\ b & = 2a - 18 &&& a & = 4 \\ b & = 2(4) - 18 \\ b & = -10 \end{align}

\begin{align} 2x + 3 & = \sqrt{2} (x + 1) \\ 2x + 3 & = x \sqrt{2} + \sqrt{2} \\ 2x - x \sqrt{2} & = \sqrt{2} - 3 \\ x(2 - \sqrt{2}) & = \sqrt{2} - 3 \\ x & = { \sqrt{2} - 3 \over 2 - \sqrt{2} } \times {2 + \sqrt{2} \over 2 + \sqrt{2}} \phantom{000000} [\text{Rationalise denominator}] \\ x & = { 2 \sqrt{2} + 2 - 6 - 3 \sqrt{2} \over (2)^2 - (\sqrt{2})^2 } \phantom{000000} [(a - b)(a + b) = a^2 - b^2] \\ x & = { -4 - \sqrt{2} \over 2} \\ x & = {-4 \over 2} - {\sqrt{2} \over 2} \\ x & = -2 - {1 \over 2} \sqrt{2} \end{align}

\begin{align} \sqrt{x - 3} + 5 & = x \\ \sqrt{x - 3} & = x - 5 \\ x - 3 & = (x - 5)^2 \\ x - 3 & = x^2 - 2(x)(5) + 5^2 \phantom{000000} [(a - b)^2 = a^2 - 2ab + b^2] \\ x - 3 & = x^2 - 10x + 25 \\ 0 & = x^2 - 11x + 28 \\ 0 & = (x - 4)(x - 7) \end{align}
\begin{align} x - 4 & = 0 && \text{ or } & x - 7 & = 0 \\ x & = 4 &&& x & = 7 \\ \\ \text{Substitute } & \text{into original eqn,} &&& \text{Substitute } & \text{into original eqn,} \\ \text{L.H.S} & = \sqrt{4 - 3} + 5 &&& \text{L.H.S} & = \sqrt{7 - 3} + 5 \\ & = 6 &&& & = 7 \\ \\ \text{R.H.S} & = 4 &&& \text{R.H.S} & = 7 \\ \\ \therefore \text{Reject } & x = 4 &&& \therefore x & = 7 \end{align}