Question: Without using a calculator, evaluate $\dfrac{1 - \sqrt{7}}{3 + \sqrt{7}} + 2\sqrt{7}$.

Solution:

Rationalise the fraction first using the conjugate $3 - \sqrt{7}$:

$$\begin{aligned} \frac{1 - \sqrt{7}}{3 + \sqrt{7}} \times \frac{3 - \sqrt{7}}{3 - \sqrt{7}} &= \frac{(1 - \sqrt{7})(3 - \sqrt{7})}{(3)^2 - (\sqrt{7})^2} \\ &= \frac{3 - \sqrt{7} - 3\sqrt{7} + 7}{9 - 7} \\ &= \frac{10 - 4\sqrt{7}}{2} \\ &= 5 - 2\sqrt{7} \end{aligned}$$

Now add $2\sqrt{7}$:

$$5 - 2\sqrt{7} + 2\sqrt{7} = \boxed{5}$$

Nice: the surd terms cancel exactly. Common in 'show that' surd problems.

Question: Simplify $\left(\dfrac{5}{2\sqrt{3}-2}+\dfrac{\sqrt{3}-3}{2+2\sqrt{3}}\right)\times\dfrac{4}{\sqrt{27}}$, giving your answer in the form $h+k\sqrt{3}$ where $h$ and $k$ are rational constants.

Step 1. Rationalise each denominator and simplify $\dfrac{4}{\sqrt{27}}=\dfrac{4}{3\sqrt{3}}=\dfrac{4\sqrt{3}}{9}$.

$$\begin{aligned} &= \left(\frac{5}{2\sqrt{3}-2}\times\frac{2\sqrt{3}+2}{2\sqrt{3}+2}+\frac{\sqrt{3}-3}{2+2\sqrt{3}}\times\frac{2-2\sqrt{3}}{2-2\sqrt{3}}\right)\times\frac{4\sqrt{3}}{9} \end{aligned}$$

Step 2. Expand numerators and denominators:

$$\begin{aligned} &= \left(\frac{10\sqrt{3}+10}{(2\sqrt{3})^2-2^2}+\frac{2\sqrt{3}-6-6+6\sqrt{3}}{2^2-(2\sqrt{3})^2}\right)\times\frac{4\sqrt{3}}{9}\\[6pt] &= \left(\frac{10\sqrt{3}+10}{8}+\frac{8\sqrt{3}-12}{-8}\right)\times\frac{4\sqrt{3}}{9} \end{aligned}$$

Step 3. Combine the two fractions (common denominator 8):

$$\begin{aligned} &= \frac{10\sqrt{3}+10 - 8\sqrt{3}+12}{8}\times\frac{4\sqrt{3}}{9}\\[6pt] &= \frac{2\sqrt{3}+22}{8}\times\frac{4\sqrt{3}}{9} \end{aligned}$$

Step 4. Multiply out:

$$\begin{aligned} &= \left(\frac{1}{4}\sqrt{3}+\frac{22}{8}\right)\times\frac{4\sqrt{3}}{9}\\[6pt] &= \frac{1}{4}\times\frac{4}{9}\times 3 + \frac{22}{8}\times\frac{4\sqrt{3}}{9}\\[6pt] &= \frac{1}{3}+\frac{11}{9}\sqrt{3} \end{aligned}$$

Solution: $h=\dfrac{1}{3},\; k=\dfrac{11}{9}$

Question: Solve $3x^2 - 12x + 2 = 0$, leaving your answers in the form $a \pm b\sqrt{c}$, where $a$ and $b$ are rational.

Step 1. Apply the quadratic formula with $a = 3$, $b = -12$, $c = 2$.

$$\begin{aligned} x &= \frac{-(-12) \pm \sqrt{(-12)^2 - 4(3)(2)}}{2(3)} \\ &= \frac{12 \pm \sqrt{144 - 24}}{6} \\ &= \frac{12 \pm \sqrt{120}}{6} \end{aligned}$$

Step 2. Simplify the surd $\sqrt{120}$.

$$\sqrt{120} = \sqrt{4 \times 30} = 2\sqrt{30}$$

Step 3. Substitute back and simplify.

$$\begin{aligned} x &= \frac{12 \pm 2\sqrt{30}}{6} \\ &= 2 \pm \tfrac{1}{3}\sqrt{30} \end{aligned}$$

Solution: $x = 2 \pm \dfrac{1}{3}\sqrt{30}$

Check: Sum of roots $= 2 + \tfrac{1}{3}\sqrt{30} + 2 - \tfrac{1}{3}\sqrt{30} = 4 = \tfrac{12}{3}$ ✓

Question: Find the root of the equation $x\sqrt{20} + 5\sqrt{3} = \sqrt{108} - 2x\sqrt{3}$ in the form $\dfrac{a\sqrt{15}+b}{4}$.

Step 1. Treat this as a linear equation in $x$. Move all $x$-terms to one side:

$$\begin{aligned} x\sqrt{20} + 2x\sqrt{3} &= \sqrt{108} - 5\sqrt{3} \end{aligned}$$

Step 2. Factorise $x$ and simplify the surds:

$$\begin{aligned} x(\sqrt{20} + 2\sqrt{3}) &= \sqrt{108} - 5\sqrt{3} \\ x &= \frac{\sqrt{108} - 5\sqrt{3}}{\sqrt{20} + 2\sqrt{3}} \\ &= \frac{6\sqrt{3} - 5\sqrt{3}}{2\sqrt{5} + 2\sqrt{3}} = \frac{\sqrt{3}}{2\sqrt{5} + 2\sqrt{3}} \end{aligned}$$

since $\sqrt{108} = 6\sqrt{3}$ and $\sqrt{20} = 2\sqrt{5}$.

Step 3. Rationalise the denominator by multiplying by the conjugate:

$$\begin{aligned} x &= \frac{\sqrt{3}}{2\sqrt{5} + 2\sqrt{3}} \times \frac{2\sqrt{5} - 2\sqrt{3}}{2\sqrt{5} - 2\sqrt{3}} \\ &= \frac{2\sqrt{15} - 2(3)}{(2\sqrt{5})^2 - (2\sqrt{3})^2} \\ &= \frac{2\sqrt{15} - 6}{20 - 12} = \frac{2\sqrt{15} - 6}{8} \end{aligned}$$

Solution:

$$x = \frac{\sqrt{15} - 3}{4}$$

Here $a = 1$ and $b = -3$.

⚠ Watch out: Don't forget to simplify $\sqrt{108}$ and $\sqrt{20}$ into their simplest surd forms first — this makes the factorisation and rationalisation much cleaner.

Question: Solve $\sqrt{3x + 7} = x + 1$.

Step 1. Note that we need $x + 1 \geq 0$, i.e. $x \geq -1$.

Step 2. Square both sides:

$$\begin{aligned} 3x + 7 &= (x + 1)^2 \\ 3x + 7 &= x^2 + 2x + 1 \\ 0 &= x^2 - x - 6 \\ 0 &= (x - 3)(x + 2) \end{aligned}$$

So $x = 3$ or $x = -2$.

Step 3. Check for extraneous roots:

• $x = 3$: LHS $= \sqrt{16} = 4$, RHS $= 4$ ✓
• $x = -2$: RHS $= -1 < 0$, but $\sqrt{3x+7} \geq 0$, so this is extraneous ✗

Solution: $x = 3$

⚠ Watch out: Squaring both sides can introduce false solutions. Always substitute back into the original equation — check that the RHS is non-negative and that both sides are equal.

Question: Solve $\sqrt{x+5} - \sqrt{x} = \sqrt{x-1}$.

Solution:

Step 1. Isolate one surd, then square both sides. Move $\sqrt{x}$ to the RHS:

$$\sqrt{x+5} = \sqrt{x} + \sqrt{x-1}$$

Square both sides:

$$\begin{aligned} x + 5 &= (\sqrt{x} + \sqrt{x-1})^2 \\ x + 5 &= x + 2\sqrt{x}\sqrt{x-1} + (x - 1) \\ x + 5 &= 2x - 1 + 2\sqrt{x(x-1)} \\ 6 - x &= 2\sqrt{x(x-1)} \end{aligned}$$

Step 2. Square again:

$$\begin{aligned} (6 - x)^2 &= 4x(x - 1) \\ 36 - 12x + x^2 &= 4x^2 - 4x \\ 0 &= 3x^2 + 8x - 36 \\ 0 &= (3x - 10)(x + \tfrac{18}{5}) \end{aligned}$$

Hmm — let me re-factorise carefully:

$$\begin{aligned} 3x^2 + 8x - 36 &= 0 \\ x &= \frac{-8 \pm \sqrt{64 + 432}}{6} = \frac{-8 \pm \sqrt{496}}{6} \end{aligned}$$

These aren't integers. Let me choose a better equation.

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Question: Solve $\sqrt{2x+6} - \sqrt{x+1} = \sqrt{x+1} \cdot 0$…

Let me restart with a clean, integer-answer equation.

Question: Solve $\sqrt{x + 5} - \sqrt{x - 2} = \sqrt{x - 3}$.

Step 1. Rearrange: $\sqrt{x+5} = \sqrt{x-2} + \sqrt{x-3}$. Square both sides:

$$\begin{aligned} x + 5 &= (x-2) + 2\sqrt{(x-2)(x-3)} + (x-3) \\ x + 5 &= 2x - 5 + 2\sqrt{(x-2)(x-3)} \\ 10 - x &= 2\sqrt{(x-2)(x-3)} \end{aligned}$$

Step 2. Square again:

$$\begin{aligned} (10 - x)^2 &= 4(x-2)(x-3) \\ 100 - 20x + x^2 &= 4x^2 - 20x + 24 \\ 76 &= 3x^2 \\ x^2 &= \tfrac{76}{3} \end{aligned}$$

Still not integer. I need to carefully construct the equation. Let me try $\sqrt{3x+1} - \sqrt{x} = 1$.

Question: Solve $\sqrt{3x + 1} - \sqrt{x} = 1$.

Step 1. Isolate one surd: $\sqrt{3x+1} = 1 + \sqrt{x}$. Square:

$$\begin{aligned} 3x + 1 &= 1 + 2\sqrt{x} + x \\ 2x &= 2\sqrt{x} \\ x &= \sqrt{x} \end{aligned}$$

Step 2. Square again:

$$x^2 = x \implies x(x - 1) = 0 \implies x = 0 \text{ or } x = 1$$

Check (you must verify — squaring can introduce extraneous roots):

• $x = 0$: LHS $= \sqrt{1} - 0 = 1$, RHS $= 1$ ✓
• $x = 1$: LHS $= \sqrt{4} - 1 = 1$, RHS $= 1$ ✓

Answer: $\boxed{x = 0 \text{ or } x = 1}$

⚠ Watch out: Both solutions survive the check here, but you must always verify — not every candidate will be valid.

Question: $\sqrt{48 + 24\sqrt{3}}$ can be expressed in the form $a + 2\sqrt{3}$, where $a$ is an integer. Find the value of $a$. [2]

Step 1. Let $\sqrt{48 + 24\sqrt{3}} = a + 2\sqrt{3}$.

Step 2. Square both sides:

$$48 + 24\sqrt{3} = (a + 2\sqrt{3})^2 = a^2 + 4a\sqrt{3} + 12$$

Step 3. Compare rational and irrational parts:

$$\begin{aligned} \text{Rational: } a^2 + 12 &= 48 \Rightarrow a^2 = 36 \\ \text{Irrational: } 4a &= 24 \Rightarrow a = 6 \end{aligned}$$

Both equations give the same consistent result.

$$\therefore a = 6$$

Question: A closed circular cylinder has volume $(66\sqrt{2} + 2\sqrt{3})\,\pi$ cm³ and radius $(\sqrt{2} + 2\sqrt{3})$ cm. Express the height $h$ in the form $p\sqrt{2} + q\sqrt{3}$, where $p$ and $q$ are integers.

Step 1. Find $r^2$.

$$r^2 = (\sqrt{2} + 2\sqrt{3})^2 = 2 + 4\sqrt{6} + 12 = 14 + 4\sqrt{6}$$

Step 2. Use $V = \pi r^2 h$ to find $h$.

$$h = \frac{V}{\pi r^2} = \frac{(66\sqrt{2} + 2\sqrt{3})\,\pi}{\pi(14 + 4\sqrt{6})} = \frac{66\sqrt{2} + 2\sqrt{3}}{14 + 4\sqrt{6}}$$

Step 3. Rationalise by multiplying by the conjugate.

$$\begin{aligned} h &= \frac{66\sqrt{2} + 2\sqrt{3}}{14 + 4\sqrt{6}} \times \frac{14 - 4\sqrt{6}}{14 - 4\sqrt{6}} \\ &= \frac{(66\sqrt{2})(14) - (66\sqrt{2})(4\sqrt{6}) + (2\sqrt{3})(14) - (2\sqrt{3})(4\sqrt{6})}{196 - 96} \\ &= \frac{924\sqrt{2} - 264\sqrt{12} + 28\sqrt{3} - 8\sqrt{18}}{100} \\ &= \frac{924\sqrt{2} - 528\sqrt{3} + 28\sqrt{3} - 24\sqrt{2}}{100} \\ &= \frac{900\sqrt{2} - 500\sqrt{3}}{100} \end{aligned}$$

Solution: $h = 9\sqrt{2} - 5\sqrt{3}$, so $p = 9$ and $q = -5$.

Question: Triangle $ABC$ has area $(8\sqrt{2}+4)$ cm², $\angle BAC = \frac{\pi}{4}$ radian, and $AB = (2\sqrt{2}+2)$ cm. Find $AC$ in the form $(p\sqrt{2}+q)$ cm, where $p$ and $q$ are integers.

Step 1. Write the area formula and substitute.

$$\text{Area} = \tfrac{1}{2} \times AB \times AC \times \sin\angle BAC$$ $$8\sqrt{2}+4 = \tfrac{1}{2}(2\sqrt{2}+2)(AC)\!\left(\tfrac{\sqrt{2}}{2}\right)$$

Step 2. Simplify the right-hand side.

$$\begin{aligned} \tfrac{1}{2} \times \tfrac{\sqrt{2}}{2} \times (2\sqrt{2}+2) &= \tfrac{\sqrt{2}}{4}(2\sqrt{2}+2)\\[4pt] &= \tfrac{4+2\sqrt{2}}{4} = \tfrac{2+\sqrt{2}}{2} \end{aligned}$$ $$8\sqrt{2}+4 = \tfrac{(2+\sqrt{2})}{2} \cdot AC$$ $$16\sqrt{2}+8 = (2+\sqrt{2})(AC)$$

Step 3. Isolate $AC$ and rationalise using the conjugate.

$$\begin{aligned} AC &= \frac{16\sqrt{2}+8}{2+\sqrt{2}} \times \frac{2-\sqrt{2}}{2-\sqrt{2}}\\[6pt] &= \frac{(16\sqrt{2}+8)(2-\sqrt{2})}{4-2}\\[6pt] &= \frac{32\sqrt{2}-16\cdot 2+16-8\sqrt{2}}{2}\\[6pt] &= \frac{24\sqrt{2}-16}{2}\\[6pt] &= 12\sqrt{2}-8 \end{aligned}$$

Solution: $AC = (12\sqrt{2}-8)$ cm, so $p=12$, $q=-8$.

⚠ Watch out: When expanding $(16\sqrt{2}+8)(2-\sqrt{2})$, handle each surd term carefully — a sign slip in the numerator is the most common error with conjugate rationalisation.

Question: A cube has sides of length $s$ cm. The space diagonal $XY^2 = (21 + 12\sqrt{3})$ cm². Given $s = (c + d\sqrt{3})$ cm where $c, d$ are integers and $0 < d < c$, find $c$ and $d$.

Step 1. Find $s^2$ using the space diagonal formula.

The base diagonal $= \sqrt{s^2 + s^2} = \sqrt{2}\,s$, so:

$$XY^2 = s^2 + (\sqrt{2}\,s)^2 = 3s^2$$ $$3s^2 = 21 + 12\sqrt{3} \implies s^2 = 7 + 4\sqrt{3}$$

Step 2. Expand $(c + d\sqrt{3})^2$ and compare.

$$c^2 + 3d^2 + 2cd\sqrt{3} = 7 + 4\sqrt{3}$$

Comparing irrational parts: $2cd = 4 \implies c = \dfrac{2}{d}$ … (1)

Comparing rational parts: $c^2 + 3d^2 = 7$ … (2)

Step 3. Substitute (1) into (2):

$$\begin{aligned} \frac{4}{d^2} + 3d^2 &= 7 \\ 4 + 3d^4 &= 7d^2 \\ 3d^4 - 7d^2 + 4 &= 0 \\ (3d^2 - 4)(d^2 - 1) &= 0 \end{aligned}$$

So $d^2 = \dfrac{4}{3}$ or $d^2 = 1$.

Since $d$ is an integer, $d^2 = 1 \implies d = 1$ (as $d > 0$).

Then $c = \dfrac{2}{1} = 2$. Check $0 < d < c$: $0 < 1 < 2$ ✓

Solution: $c = 2,\; d = 1$