Algebra Overview – A Level Maths Year 1

1 The Big Idea

Algebra and functions is the largest topic in the first year of A Level, and the toolkit for everything else. Differentiation, integration, trigonometry, logarithms and even mechanics all end in an equation or an expression that has to be simplified, factorised or solved — and those skills live here.

At GCSE you met most of these ideas one at a time. At A Level you are expected to choose the right tool without being told, to work exactly (surds and fractions rather than decimals), and to connect every piece of algebra to a graph: a root is a crossing point, a discriminant counts intersections, an inequality is a region.

What is new compared with Higher GCSE

Fractional and negative indices in context · the discriminant
Quadratics in disguise · set notation for inequalities
The factor theorem · cubics, quartics and reciprocal graphs
Transformations including $y = \mathrm{f}(ax)$ · the binomial expansion

Calculators are allowed in A Level papers, but many algebra questions say "show detailed reasoning" or ask for an exact answer. Then the algebraic method must be written in full; an answer read off a calculator's equation solver scores nothing.

2 The Language You Need

Word	Meaning	Example
Surd	An irrational root, kept exact.	$\sqrt{5}$, $3 - 2\sqrt{2}$
Rationalise	Remove surds from a denominator.	$\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}$
Identity	True for every value; written $\equiv$.	$x^2 - 4 \equiv (x-2)(x+2)$
Root	A solution of $\mathrm{f}(x) = 0$; where $y = \mathrm{f}(x)$ meets the $x$-axis.	$x = 3$ is a root of $x^2 - 9$
Discriminant	$b^2 - 4ac$ — decides how many real roots a quadratic has.	$x^2 + x + 1$: $\;1 - 4 = -3$
Vertex	The turning point of a parabola.	$(x-2)^2 + 5$ has vertex $(2, 5)$
Degree	The highest power in a polynomial.	$x^4 - x$ has degree $4$
Asymptote	A line a curve approaches but never reaches.	$y = \dfrac{1}{x}$: both axes
$\mathbf{f}(x)$	Function notation — the output when the input is $x$.	$\mathrm{f}(x) = x^3 - 2x$
$\propto$	"Is proportional to".	$y \propto x^2$ means $y = kx^2$

Set notation for solutions

$\{x : x < -3\} \cup \{x : x > 4\}$ means "$x < -3$ or $x > 4$"
$\{x : x > 1\} \cap \{x : x < 5\}$ means "$1 < x < 5$"

"Or" is not "and". Two separate intervals must be joined with "or" (or $\cup$). Writing $-3 > x > 4$ is meaningless — no number is both less than $-3$ and greater than $4$.

3 The Core Techniques

1 — Indices

$a^{-n} = \dfrac{1}{a^n}$, $a^{\frac{m}{n}} = \left(\sqrt[n]{a}\right)^m$

2 — Completing the square

$ax^2 + bx + c \equiv a(x + p)^2 + q$, vertex at $(-p,\, q)$

3 — The discriminant

$b^2 - 4ac > 0$: two real roots · $= 0$: one repeated root · $< 0$: no real roots

4 — The factor theorem

$\mathrm{f}(a) = 0 \iff (x - a)$ is a factor of $\mathrm{f}(x)$

5 — Transformations

$y = \mathrm{f}(x + a)$: left by $a$ · $y = \mathrm{f}(x) + a$: up by $a$ · $y = a\mathrm{f}(x)$: stretch $\times a$ vertically · $y = \mathrm{f}(ax)$: stretch $\times \tfrac{1}{a}$ horizontally

6 — The binomial expansion

$(a + b)^n = \displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$ for positive integer $n$

Worked Example — Several techniques in one question

The line $y = 2x + c$ and the curve $y = x^2 + 3$ are drawn on the same axes. (a) Find the value of $c$ for which the line is a tangent to the curve, and the point of contact. (b) For what values of $c$ does the line meet the curve at two points?

①At an intersection, $x^2 + 3 = 2x + c$, so $x^2 - 2x + (3 - c) = 0$.

②Discriminant: $(-2)^2 - 4(1)(3 - c) = 4c - 8$.

③Tangent means exactly one meeting point, so $4c - 8 = 0$ and $c = 2$.

④Then $x^2 - 2x + 1 = (x - 1)^2 = 0$, so $x = 1$ and $y = 4$. Point of contact $(1, 4)$.

⑤Two points need $4c - 8 > 0$, so $c > 2$.

One question, four skills: simultaneous equations, the discriminant, factorising and a linear inequality. A Level questions are built this way — expect to link topics.

4 The Algebra and Functions Subtopics

The pages in this topic. The codes are the DfE content areas: B Algebra and functions, D Sequences and series.

BIndices and SurdsLaws of indices including negative and fractional powers; manipulating surds and rationalising denominators.
BQuadratic FunctionsGraphs, the discriminant, completing the square, and quadratics in a function of $x$.
BSimultaneous EquationsElimination and substitution, one linear and one quadratic, and what the solutions mean on a graph.
BLinear and Quadratic InequalitiesSolving and writing answers in set notation; shading regions with solid and dashed boundaries.
BPolynomials and the Factor TheoremExpanding, factorising, simple algebraic division and the factor theorem.
BCurve Sketching and ProportionCubics, quartics, $y = \dfrac{a}{x}$ and $y = \dfrac{a}{x^2}$, asymptotes, intersections, and direct and inverse proportion.
BTransformations of Graphs$y = a\mathrm{f}(x)$, $y = \mathrm{f}(x) + a$, $y = \mathrm{f}(x + a)$, $y = \mathrm{f}(ax)$ and combinations.
DThe Binomial Expansion$(a + bx)^n$ for positive integer $n$, $n!$ and $^n\mathrm{C}_r$, Pascal's triangle and approximations.

5 Common Mistakes to Avoid

Negative index means reciprocal, not negative. $2^{-3} = \dfrac{1}{8}$, not $-8$.

Inventing surd rules. $\sqrt{a + b} \neq \sqrt{a} + \sqrt{b}$. Only multiplication and division split across a root.

Treating a quadratic inequality like an equation. $x^2 > 9$ gives $x < -3$ or $x > 3$, not just $x > 3$. Always sketch.

Getting horizontal transformations backwards. $y = \mathrm{f}(x + 3)$ moves the graph left by $3$, and $y = \mathrm{f}(2x)$ squashes it by a factor of $\tfrac{1}{2}$. Changes inside the bracket act on $x$ "the opposite way".

Dropping brackets in the binomial expansion. In $(1 - 2x)^6$ the second term of the bracket is $-2x$, and $(-2x)^2 = 4x^2$.

Dividing by an expression that could be zero. Never cancel $x$ from both sides of $x^3 = 4x$ — you lose the root $x = 0$. Factorise instead: $x(x - 2)(x + 2) = 0$.

6 Quick Reference

Fractional index

$a^{\frac{m}{n}}$: root $n$, then power $m$.

Rationalising

Multiply by the conjugate: $a + \sqrt{b} \to a - \sqrt{b}$.

Discriminant

$b^2 - 4ac$: $>0$ two roots, $=0$ one, $<0$ none.

Tangency

Line meets curve once $\iff$ discriminant $= 0$.

Inequalities

Solve $= 0$, sketch, read the region; reverse for negative multipliers.

Region boundaries

Solid for $\leqslant, \geqslant$; dashed for $<, >$.

Factor theorem

$\mathrm{f}(a) = 0 \iff (x - a)$ is a factor.

Repeated root

Graph touches the axis there.

Proportion

$y \propto x^n$: $y = kx^n$; $y \propto \dfrac{1}{x}$: $y = \dfrac{k}{x}$.

Binomial

General term $\dbinom{n}{r}a^{n-r}b^r$.

7 Practice Questions

Ten questions drawing on the whole Algebra and Functions topic. Work each one through before opening the solution.

Question 1

Evaluate $125^{-\frac{2}{3}}$ without a calculator.

▶ Show solution

$\sqrt[3]{125} = 5$, and $5^2 = 25$.

The negative index gives the reciprocal: $125^{-\frac23} = \dfrac{1}{25}$.

Question 2

Express $\dfrac{6}{\sqrt{5} - 1}$ in the form $a + b\sqrt{5}$, where $a$ and $b$ are rational.

▶ Show solution

Multiply top and bottom by $\sqrt{5} + 1$. Denominator: $5 - 1 = 4$.

$\dfrac{6\left(\sqrt{5} + 1\right)}{4} = \dfrac{3}{2} + \dfrac{3}{2}\sqrt{5}$

So $a = \tfrac32$, $b = \tfrac32$.

Question 3

Express $2x^2 - 12x + 23$ in the form $a(x + p)^2 + q$, and hence state the coordinates of the vertex of $y = 2x^2 - 12x + 23$.

▶ Show solution

Take out the $2$ from the $x$ terms: $2\left(x^2 - 6x\right) + 23$.

$x^2 - 6x = (x - 3)^2 - 9$, so $2\left[(x-3)^2 - 9\right] + 23 = 2(x - 3)^2 - 18 + 23$.

$2x^2 - 12x + 23 \equiv 2(x - 3)^2 + 5$

The vertex (a minimum, as $a > 0$) is at $(3, 5)$.

Question 4

The equation $x^2 + kx + 9 = 0$ has equal roots. Find the possible values of $k$.

▶ Show solution

Equal roots means the discriminant is zero: $k^2 - 4(1)(9) = 0$.

$k^2 = 36$, so $k = 6$ or $k = -6$.

(Check: $x^2 + 6x + 9 = (x + 3)^2$ and $x^2 - 6x + 9 = (x - 3)^2$ ✓)

Question 5

Solve the simultaneous equations $y = x + 1$ and $x^2 + y^2 = 13$.

▶ Show solution

Substitute: $x^2 + (x + 1)^2 = 13$, so $2x^2 + 2x + 1 = 13$.

$2x^2 + 2x - 12 = 0 \;\Rightarrow\; x^2 + x - 6 = 0 \;\Rightarrow\; (x + 3)(x - 2) = 0$

$x = 2,\; y = 3$ or $x = -3,\; y = -2$.

Geometrically, the line cuts the circle at two points.

Question 6

Solve $x^2 - x - 12 > 0$, giving your answer in set notation.

▶ Show solution

$x^2 - x - 12 = (x - 4)(x + 3)$, with roots $x = -3$ and $x = 4$.

A positive quadratic is above the axis outside the roots.

$\{x : x < -3\} \cup \{x : x > 4\}$

Question 7

$\mathrm{f}(x) = x^3 - 7x - 6$. Show that $(x + 1)$ is a factor of $\mathrm{f}(x)$ and hence factorise $\mathrm{f}(x)$ completely.

▶ Show solution

$\mathrm{f}(-1) = -1 + 7 - 6 = 0$, so $(x + 1)$ is a factor by the factor theorem.

Dividing: $x^3 - 7x - 6 = (x + 1)\left(x^2 - x - 6\right)$.

$x^2 - x - 6 = (x - 3)(x + 2)$, so $\mathrm{f}(x) = (x + 1)(x + 2)(x - 3)$.

Question 8

The curve $y = \mathrm{f}(x)$ has a maximum point at $(2, 5)$. Write down the coordinates of the maximum point on each of: (a) $y = \mathrm{f}(x - 3)$ (b) $y = 2\mathrm{f}(x)$ (c) $y = \mathrm{f}(2x)$ (d) $y = \mathrm{f}(x) - 4$.

▶ Show solution

(a) Translation $3$ to the right: $(5, 5)$.

(b) Vertical stretch, factor $2$: $(2, 10)$.

(d) Translation $4$ down: $(2, 1)$.

Question 9

Find the coefficient of $x^3$ in the expansion of $(1 - 2x)^8$.

▶ Show solution

$\dbinom{8}{3}(-2x)^3 = 56 \times (-8x^3)$

Coefficient $= -448$.

Question 10

The intensity $I$ of light from a lamp is inversely proportional to the square of the distance $d$ metres from the lamp. When $d = 3$, $I = 20$.

(a) Find a formula for $I$ in terms of $d$. (b) Find $I$ when $d = 6$. (c) Find $d$ when $I = 45$. (d) Describe the shape of the graph of $I$ against $d$ for $d > 0$.

▶ Show solution

(a) $I = \dfrac{k}{d^2}$. Then $20 = \dfrac{k}{9}$, so $k = 180$ and $I = \dfrac{180}{d^2}$.

(b) $I = \dfrac{180}{36} = 5$. (Doubling the distance quarters the intensity.)

(c) $45 = \dfrac{180}{d^2}$, so $d^2 = 4$ and $d = 2$ (distance is positive).

(d) A $y = \dfrac{a}{x^2}$ curve in the first quadrant: decreasing, with the $I$-axis ($d = 0$) and the $d$-axis ($I = 0$) as asymptotes.

Algebra Overview · A Level Maths Year 1 · Created with MathJax