Surds

The largest perfect-square factor of 72 is 36, and $72 = 36 \times 2$, so $\sqrt{72} = \sqrt{36} \times \sqrt{2} = 6\sqrt{2}$. ($2\sqrt{18}$ used the smaller factor 4; $\sqrt{18}$ still simplifies to $3\sqrt{2}$, giving $6\sqrt{2}$ anyway. 36 forgot the $\sqrt{2}$.)

The conjugate creates a difference of two squares, $a^2 - b^2$, which removes the surd and leaves a rational number, e.g. $(2 + \sqrt{3})(2 - \sqrt{3}) = 4 - 3 = 1$. This is exactly why conjugates are used to rationalise denominators.

Simplify each first: $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{32} = 4\sqrt{2}$, now like surds: $2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2}$. ($\sqrt{40}$ added the radicands without simplifying; $2\sqrt{2} + 4\sqrt{2}$ is the unsimplified intermediate.)

Multiply rule: $\sqrt{6} \times \sqrt{10} = \sqrt{60}$, then $\sqrt{60} = \sqrt{4 \times 15} = 2\sqrt{15}$. ($\sqrt{60}$ is unsimplified; $\sqrt{16}$ added the radicands instead of multiplying.)

Divide rule: $\dfrac{\sqrt{48}}{\sqrt{3}} = \sqrt{\dfrac{48}{3}} = \sqrt{16} = 4$. ($\sqrt{45}$ subtracted the radicands; $\sqrt{16}$ and 16 did not take the final square root.)

$\sqrt{2}$ and $\sqrt{3}$ are unlike surds, so they cannot be combined into a single surd, and their sum is irrational. In particular $\sqrt{2} + \sqrt{3} \ne \sqrt{5}$ (you cannot add the numbers under the roots).

Square both the coefficient and the surd: $(2\sqrt{3})^2 = 2^2 \times (\sqrt{3})^2 = 4 \times 3 = 12$. (6 did not square either; $4\sqrt{3}$ squared only the coefficient.)

Distribute: $\sqrt{2} \times \sqrt{6} = \sqrt{12} = 2\sqrt{3}$ and $\sqrt{2} \times \sqrt{2} = 2$, so the result is $2\sqrt{3} - 2$. ($\sqrt{12} - 2$ did not simplify $\sqrt{12}$; $2 - 2\sqrt{3}$ swapped the signs.)

FOIL: $(\sqrt{2})(\sqrt{2}) + (\sqrt{2})(1) + (3)(\sqrt{2}) + (3)(1) = 2 + \sqrt{2} + 3\sqrt{2} + 3 = 5 + 4\sqrt{2}$. ($3 + 4\sqrt{2}$ added the rationals as $2 + 1$; $6 + 4\sqrt{2}$ added them as $3 + 3$.)

Difference of squares: $(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$ (always rational). (8 added instead of subtracting; $2\sqrt{15}$ kept the middle terms, but they cancel.)

Multiply top and bottom by $\sqrt{2}$: $\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$. ($\dfrac{1}{2}$ squared the denominator but not the numerator; $\dfrac{2}{\sqrt{2}}$ still has a surd in the denominator.)

$\dfrac{5}{\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{5\sqrt{3}}{3}$. ($\dfrac{5}{3}$ forgot the numerator surd; $\dfrac{\sqrt{3}}{5}$ inverted the fraction.)

The diagonal splits the square into two right-angled triangles, so by Pythagoras $d^2 = 5^2 + 5^2 = 50$ and $d = \sqrt{50} = 5\sqrt{2}$ cm. ($10$ cm forgot to take the square root.)

Using the distance formula, $d = \sqrt{(5 - 1)^2 + (7 - 1)^2} = \sqrt{16 + 36} = \sqrt{52}$. Then $\sqrt{52} = \sqrt{4 \times 13} = 2\sqrt{13}$. ($\sqrt{52}$ is correct but not simplified.)

Multiply by the conjugate $1 - \sqrt{2}$: $\dfrac{1 - \sqrt{2}}{(1 + \sqrt{2})(1 - \sqrt{2})} = \dfrac{1 - \sqrt{2}}{1 - 2} = \dfrac{1 - \sqrt{2}}{-1} = \sqrt{2} - 1$. ($\dfrac{1 + \sqrt{2}}{3}$ squared the denominator $(1 + 2)$ instead of using difference of squares; $1 - \sqrt{2}$ forgot to divide by $-1$.)

Use $(a + b)^2 = a^2 + 2ab + b^2$ with $a = \sqrt{3}$, $b = \sqrt{2}$: $3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$. (5 forgot the middle term, the canonical error; $5 + 2\sqrt{5}$ added the radicands when multiplying $\sqrt{3} \times \sqrt{2}$.)

In a 30-60-90 triangle the sides are in the ratio $1 : \sqrt{3} : 2$ (opposite $30^\circ$, $60^\circ$, $90^\circ$). The side opposite $30^\circ$ is $4$, so the side opposite $60^\circ$ is $4\sqrt{3}$ cm. ($8$ is the hypotenuse, opposite $90^\circ$.)

Area $= (2 + \sqrt{3})^2 = 2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$ cm². ($7$ forgot the middle term $4\sqrt{3}$, and $4 + 4\sqrt{3}$ forgot to add the $(\sqrt{3})^2 = 3$.)

By Pythagoras, $(\sqrt{41})^2 = 4^2 + b^2$, so $b^2 = 41 - 16 = 25$ and $b = 5$ cm. ($\sqrt{57}$ added instead of subtracting.)

Multiply by the conjugate $\sqrt{5} + 2$: $\dfrac{\sqrt{5} + 2}{(\sqrt{5})^2 - 2^2} = \dfrac{\sqrt{5} + 2}{5 - 4} = \sqrt{5} + 2$. ($\dfrac{\sqrt{5} + 2}{9}$ used $(\sqrt{5})^2 + 2^2 = 9$, but difference of squares uses minus.)

Simplify the surds: $\sqrt{8} = 2\sqrt{2}$ and $\sqrt{18} = 3\sqrt{2}$, so $\sqrt{8} + \sqrt{18} = 5\sqrt{2}$, then $\dfrac{5\sqrt{2}}{\sqrt{2}} = 5$. ($\sqrt{13}$ added the radicands; $5\sqrt{2}$ did not divide by $\sqrt{2}$.)

Area $= \sqrt{20} \times \sqrt{5} = \sqrt{20 \times 5} = \sqrt{100} = 10$ cm². (Multiplying two surds can give a whole number; $100$ cm² forgot to take the square root.)

$x = \sqrt{50} = \sqrt{25 \times 2} = 5\sqrt{2}$ (the positive root). ($\sqrt{50}$ is unsimplified; 25 squared instead of taking the root; $\pm 5\sqrt{2}$ gave both signs when only the positive was asked.)

Use $(a - b)^2 = a^2 - 2ab + b^2$ with $a = 3\sqrt{2}$, $b = \sqrt{5}$: $(3\sqrt{2})^2 - 2(3\sqrt{2})(\sqrt{5}) + (\sqrt{5})^2 = 18 - 6\sqrt{10} + 5 = 23 - 6\sqrt{10}$. ($23 + 6\sqrt{10}$ forgot the minus on the middle term; $13 - 6\sqrt{10}$ subtracted 5 instead of adding (since $(-\sqrt{5})^2 = +5$); $23 - 2\sqrt{10}$ forgot the factor of 3.)

Multiply top and bottom by the conjugate $\sqrt{3} + 1$. Denominator: $(\sqrt{3} - 1)(\sqrt{3} + 1) = 3 - 1 = 2$. Numerator: $(\sqrt{3} + 1)^2 = 3 + 2\sqrt{3} + 1 = 4 + 2\sqrt{3}$. So $\dfrac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$. ($\dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$ used the wrong sign in the numerator square.)