The three you must know cold
$\sqrt{72} = \sqrt{36 \times 2} = 6\sqrt{2}$
Rationalise $\dfrac{6}{3+\sqrt{5}}$
- Multiply top and bottom by the conjugate: $3-\sqrt{5}$
- $\dfrac{6(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$
- Denominator: $9 - 5 = 4$
- $\dfrac{18 - 6\sqrt{5}}{4} = \dfrac{9 - 3\sqrt{5}}{2}$
Show that $\dfrac{(2+\sqrt{3})^2}{\sqrt{12}}$ can be written as $a + b\sqrt{3}$
Numerator: $(2+\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$
Denominator: $\sqrt{12} = 2\sqrt{3}$
Expression: $\dfrac{7 + 4\sqrt{3}}{2\sqrt{3}}$. Rationalise by multiplying by $\dfrac{\sqrt{3}}{\sqrt{3}}$:
$\dfrac{(7+4\sqrt{3})\sqrt{3}}{2\sqrt{3}\cdot\sqrt{3}} = \dfrac{7\sqrt{3}+12}{6} = 2 + \dfrac{7\sqrt{3}}{6}$
So $a = 2$, $b = \dfrac{7}{6}$.
Memorise these six
- $a^m \times a^n = a^{m+n}$
- $a^m \div a^n = a^{m-n}$
- $(a^m)^n = a^{mn}$
- $a^0 = 1$
- $a^{-n} = \dfrac{1}{a^n}$
- $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m$
$8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4$
Evaluate $\left(\dfrac{25}{16}\right)^{-3/2}$
- Negative power → flip the fraction: $\left(\dfrac{16}{25}\right)^{3/2}$
- Apply the fractional power to top and bottom:
- $\dfrac{16^{3/2}}{25^{3/2}} = \dfrac{(\sqrt{16})^3}{(\sqrt{25})^3} = \dfrac{4^3}{5^3}$
- $= \dfrac{64}{125}$
Given $2^x \times 4^{x+1} = 8^{x-2}$, find $x$.
Get a common base. Everything is a power of 2:
$2^x \times (2^2)^{x+1} = (2^3)^{x-2}$
$2^x \times 2^{2x+2} = 2^{3x-6}$
$2^{3x+2} = 2^{3x-6}$
$3x + 2 = 3x - 6$ → $2 = -6$. No solution. (Trap question — be confident saying so.)
Always factorise first
- Factorise every numerator and denominator fully.
- Cancel common brackets.
- For + or −: find a common denominator (multiply the brackets).
- For × or ÷: flip if dividing, then cancel diagonally.
• Common factor first
• Difference of two squares: $a^2 - b^2 = (a+b)(a-b)$
• Quadratics $ax^2+bx+c$
Simplify $\dfrac{x^2 - 9}{x^2 + 5x + 6}$
Top: difference of squares → $(x+3)(x-3)$
Bottom: factorise → $(x+2)(x+3)$
$\dfrac{(x+3)(x-3)}{(x+2)(x+3)} = \dfrac{x-3}{x+2}$
Simplify $\dfrac{2x^2+7x+3}{x^2-9} \div \dfrac{2x+1}{x-3}$
Factorise:
$2x^2 + 7x + 3 = (2x+1)(x+3)$ $x^2 - 9 = (x+3)(x-3)$
Flip the second fraction (dividing):
$\dfrac{(2x+1)(x+3)}{(x+3)(x-3)} \times \dfrac{x-3}{2x+1}$
Cancel: $(2x+1)$, $(x+3)$, and $(x-3)$ all cancel.
Answer: $\mathbf{1}$
Complete the square for $x^2 + bx + c$
- Halve $b$: call it $p = b/2$
- Write $(x + p)^2$
- Subtract $p^2$ (because $(x+p)^2$ has an extra $p^2$)
- Add the original $c$
• Minimum value $= k$
• Turning point at $(h, k)$
• Line of symmetry $x = h$
Complete the square: $x^2 - 6x + 11$
- Half of $-6$ is $-3$
- $(x - 3)^2 = x^2 - 6x + 9$
- We need $+11$, but have $+9$, so add $+2$
- $x^2 - 6x + 11 = (x-3)^2 + 2$
So: minimum value is $2$, at $x = 3$. The graph never touches the x-axis (always $\geq 2$).
Write $2x^2 + 12x - 5$ in the form $a(x+b)^2 + c$
- Factor out the 2 from the $x$ terms only: $2(x^2 + 6x) - 5$
- Complete square inside: $x^2 + 6x = (x+3)^2 - 9$
- Substitute: $2[(x+3)^2 - 9] - 5$
- Expand: $2(x+3)^2 - 18 - 5 = 2(x+3)^2 - 23$
So $a = 2$, $b = 3$, $c = -23$. Turning point: $(-3, -23)$.
Solve $x^2 - 2x - 8 > 0$
Factorise: $(x-4)(x+2) > 0$
Critical values: $x = 4$ and $x = -2$
Sketch a positive parabola crossing at $-2$ and $4$. We want where it's above the x-axis.
Answer: $x < -2$ or $x > 4$
Substitution every time
- Rearrange the linear equation for one variable.
- Substitute into the quadratic.
- Expand, simplify, set $= 0$.
- Solve the resulting quadratic.
- Substitute each $x$ back to find its $y$ — pair them up.
Solve: $y = x + 1$ and $x^2 + y^2 = 25$
Sub $y = x+1$ into the second:
$x^2 + (x+1)^2 = 25$
$x^2 + x^2 + 2x + 1 = 25$
$2x^2 + 2x - 24 = 0$
$x^2 + x - 12 = 0$
$(x+4)(x-3) = 0$ → $x = -4$ or $x = 3$
Pair up:
If $x = -4$, $y = -3$. If $x = 3$, $y = 4$.
Three things to remember
- Travel routes: $\overrightarrow{AB} = -\overrightarrow{BA}$. To go from A to B, go via any point in between.
- Parallel vectors: one is a scalar multiple of the other. $\mathbf{a}$ and $3\mathbf{a}$ are parallel.
- Collinear (straight line): if $\overrightarrow{XY} = k \cdot \overrightarrow{YZ}$ (sharing point Y) then X, Y, Z are on a straight line.
OABC is a parallelogram. $\overrightarrow{OA} = \mathbf{a}$, $\overrightarrow{OC} = \mathbf{c}$. M is the midpoint of AB. Find $\overrightarrow{OM}$.
Since OABC is a parallelogram, $\overrightarrow{AB} = \overrightarrow{OC} = \mathbf{c}$.
M is the midpoint of AB, so $\overrightarrow{AM} = \dfrac{1}{2}\mathbf{c}$.
Travel O → A → M:
$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \mathbf{a} + \dfrac{1}{2}\mathbf{c}$
Strategy
- Find both vectors using the routes given (in terms of $\mathbf{a}$ and $\mathbf{b}$).
- Factorise each so it looks like $k(\text{something})$.
- Show one is a multiple of the other — they're parallel.
- State they share a common point → therefore collinear (on the same straight line).
Example phrasing for full marks:
"$\overrightarrow{XY} = 2\mathbf{a} + 4\mathbf{b} = 2(\mathbf{a} + 2\mathbf{b})$ and $\overrightarrow{YZ} = 3\mathbf{a} + 6\mathbf{b} = 3(\mathbf{a} + 2\mathbf{b})$. Since $\overrightarrow{YZ} = \dfrac{3}{2}\overrightarrow{XY}$, the vectors are parallel. As they share point Y, the points X, Y, Z are collinear."
Examiners want these phrases verbatim
- Angle at centre is twice angle at circumference (same arc).
- Angle in a semicircle is 90° (special case of above).
- Angles in the same segment are equal (subtended by same arc).
- Opposite angles in a cyclic quadrilateral sum to 180°.
- A tangent meets a radius at 90°.
- Two tangents from an external point are equal in length.
- Perpendicular from the centre bisects a chord.
- Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
How to attack any circle theorem problem
- Mark every given angle on the diagram.
- Look for: radii (→ isosceles triangles), tangents (→ 90°), diameters (→ 90° at circumference).
- Find one new angle at a time, writing the theorem name next to each step.
- Don't skip steps — every angle needs a justification.
The non-negotiable table
| θ | 0° | 30° | 45° | 60° | 90° |
|---|---|---|---|---|---|
| sin θ | 0 | $\frac{1}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{\sqrt{3}}{2}$ | 1 |
| cos θ | 1 | $\frac{\sqrt{3}}{2}$ | $\frac{\sqrt{2}}{2}$ | $\frac{1}{2}$ | 0 |
| tan θ | 0 | $\frac{1}{\sqrt{3}}$ | 1 | $\sqrt{3}$ | undef. |
Triangle ABC has $\angle B = 90°$, $\angle A = 30°$, hypotenuse $AC = 8$. Find BC exactly.
BC is opposite to the $30°$ angle. $\sin 30° = \dfrac{BC}{8}$
$\dfrac{1}{2} = \dfrac{BC}{8}$ → $BC = 4$
Now find AB: $\cos 30° = \dfrac{AB}{8}$ → $AB = 8 \times \dfrac{\sqrt{3}}{2} = 4\sqrt{3}$
Finding the bounds of a result
- Addition: UB = UB + UB | LB = LB + LB
- Subtraction: UB = UB − LB | LB = LB − UB
- Multiplication: UB = UB × UB | LB = LB × LB
- Division: UB = UB ÷ LB | LB = LB ÷ UB
Bound = value ± (half the rounding unit).
$a = 6.3$ (1 dp), $b = 2.5$ (1 dp). Find UB of $\dfrac{a}{b}$.
$a$: $6.25 \leq a < 6.35$
$b$: $2.45 \leq b < 2.55$
UB of $\dfrac{a}{b}$ = $\dfrac{\text{UB of }a}{\text{LB of }b} = \dfrac{6.35}{2.45}$
$= \dfrac{635}{245} = \dfrac{127}{49} \approx 2.59$
The algebra trick
- Let $x = $ the recurring decimal.
- Multiply by $10^n$ where $n$ = number of recurring digits, to shift one full cycle.
- Subtract: the recurring parts cancel.
- Solve for $x$ as a fraction. Simplify.
Express $0.\dot{2}\dot{7}$ as a fraction
Let $x = 0.272727\ldots$
$100x = 27.272727\ldots$
Subtract: $99x = 27$
$x = \dfrac{27}{99} = \dfrac{3}{11}$
Express $0.41\dot{6}$ as a fraction
Only the 6 recurs. Let $x = 0.41666\ldots$
$10x = 4.1666\ldots$ $100x = 41.666\ldots$
Subtract: $100x - 10x = 41.666\ldots - 4.1666\ldots$
$90x = 37.5$ → $x = \dfrac{37.5}{90} = \dfrac{75}{180} = \dfrac{5}{12}$
Inverse & composite
- $f(x)$ — apply $f$ to $x$.
- $fg(x)$ — apply $g$ FIRST, then $f$. (Read right to left.)
- $f^{-1}(x)$ — inverse: undoes $f$. To find it: set $y = f(x)$, swap $x$ and $y$, solve for $y$.
Example: $f(x) = 3x - 5$. Find $f^{-1}(x)$.
$y = 3x - 5$ → swap: $x = 3y - 5$ → $y = \dfrac{x+5}{3}$
So $f^{-1}(x) = \dfrac{x+5}{3}$.
Outside the bracket vs inside
- $y = f(x) + a$ — shift UP by $a$
- $y = f(x + a)$ — shift LEFT by $a$ (yes, left!)
- $y = -f(x)$ — reflect in the $x$-axis
- $y = f(-x)$ — reflect in the $y$-axis
- $y = af(x)$ — stretch vertically, factor $a$
- $y = f(ax)$ — stretch horizontally, factor $\frac{1}{a}$
Memorise These — Not Given in Paper 1
Mixed Quiz — Click to reveal answers
1 minute per mark
80 marks in 90 minutes leaves you 10 minutes for checking. If a 3-mark question takes more than 4 minutes, move on — come back later.
Marks for method
Even with a wrong final answer, you get method marks. A blank page gets zero. Always write a first line of working — at least define variables or rewrite the question algebraically.
Sense-check answers
Is the side of a triangle negative? Probably wrong. Is the probability over 1? Wrong. Did you answer the actual question (e.g. "find $x+y$" not just $x$)?
The target answer is given
Every line must lead clearly to the stated result. Don't skip steps. Don't write the answer at the start — work toward it. The examiner is checking your reasoning, not the final value (which is already there).
Algebraic proof skeleton
- Let the integer be $n$ (or two integers $n$ and $m$).
- Write the expression using $n$.
- Expand and simplify.
- Factorise to show it's even / odd / a multiple of something.
- State your conclusion clearly: "Therefore..."