Quadratics

CIE A-Level Mathematics· 21 min read

1. Forms of Quadratic Expressions★☆☆☆☆⏱ 4 min

📘 Definition

Quadratic Expression

ax^2 + bx + c, \quad a \neq 0

A second-degree polynomial in one variable, where $a, b, c$ are constants and the leading coefficient $a$ is non-zero.

Example:

$3 x^{2} - 5 x + 2$ is quadratic; $4 x + 7$ is linear, so not quadratic.

Quadratics can be written in three useful forms, each suited to a different purpose:

General form: $a x^{2} + b x + c$ , used for expanding and calculating discriminant
Factorised form: $a (x - p) (x - q)$ , used for identifying roots and solving equations
Completed square form: $a (x - h)^{2} + k$ , used for finding the vertex (turning point) of the parabola

📐 Worked Example

Rewrite $2 x^{2} - 8 x + 3$ in completed square form

1
Factor out the leading coefficient from the first two terms:
$2\left(x^2 - 4x\right) + 3$
2
Complete the square inside the bracket by adding and subtracting the square of half the coefficient of $x$ :
$2\left[(x^2 -4x + 4) -4\right] + 3$
3
Simplify by expanding the constant term:
$2(x-2)^2 - 8 + 3 = 2(x-2)^2 - 5$
4
Read the turning point straight from the completed square form $a (x - h)^{2} + k$ , whose vertex is at $(h, k)$ :
$(2, -5)$

2. Solving Quadratic Equations★★☆☆☆⏱ 5 min

A quadratic equation has the form $a x^{2} + b x + c = 0$ with $a \neq = 0$ . There are three standard methods for solving, each with different use cases.

Methods compared

Choose your method based on the question and form of the quadratic:

Factorisation

Rearrange to $a x^{2} + b x + c = 0$ , factorise into two linear brackets, set each bracket equal to zero

+ Pros: Fast for simple quadratics with integer roots

− Cons: Only works for factorisable quadratics

Quadratic Formula

Substitute $a, b, c$ into the formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$

+ Pros: Works for all quadratic equations

− Cons: Easy to make sign errors when substituting

Completing the Square

Rearrange to completed square form, isolate the squared term, then take square roots of both sides

+ Pros: Useful for non-calculator surd answers

− Cons: Requires more algebraic steps than other methods

📐 Worked Example

Solve $2 x^{2} - 5 x - 3 = 0$ using the quadratic formula

1
Identify $a, b, c$ with their correct signs:
$a=2, \quad b=-5, \quad c=-3$
2
Substitute into the quadratic formula:
$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-3)}}{2(2)}$
3
Simplify the expression under the square root:
$x = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm 7}{4}$
4
Calculate both roots:
$x = 3 \quad \text{or} \quad x = -\frac{1}{2}$

3. Discriminant and Nature of Roots★★☆☆☆⏱ 3 min

📘 Definition

Discriminant

\Delta = b^2 - 4ac

A value calculated from the coefficients of a quadratic that tells us the number and nature of the real roots.

Example:

For $x^{2} - 3 x + 2$ , $Δ = 9 - 8 = 1$

$Δ > 0$ : Two distinct real roots
$Δ = 0$ : One repeated (equal) real root
$Δ < 0$ : No real roots (not required for P1 to go further)

📐 Worked Example

Find the range of $k$ for which $x^{2} + 2 k x + 9 = 0$ has two distinct real roots

1
For two distinct real roots, we need $Δ > 0$ :
$\Delta = (2k)^2 - 4(1)(9) = 4k^2 - 36$
2
Set the discriminant greater than zero and simplify:
$4k^2 - 36 > 0 \implies k^2 - 9 > 0 \implies (k-3)(k+3) > 0$
3
Solve the inequality to get the final range:
$k < -3 \quad \text{or} \quad k > 3$

📐 Worked Example

Find the value of $k$ for which $y = x + k$ is a tangent to the curve $y = x^{2} + 5 x + 2$

1
A point common to the line and curve satisfies both equations, so set the two expressions for $y$ equal and rearrange into a single quadratic:
$x^2 + 5x + 2 = x + k \implies x^2 + 4x + (2 - k) = 0$
2
A tangent touches the curve at exactly one point, so the quadratic has a repeated root, which means $Δ = 0$ :
$\Delta = 4^2 - 4(1)(2 - k) = 0$
3
Expand and solve for $k$ :
$16 - 8 + 4k = 0 \implies 4k = -8 \implies k = -2$

4. Solving Quadratic Inequalities★★★☆☆⏱ 4 min

To solve a quadratic inequality, follow this structured method:

Rearrange the inequality so all terms are on one side, with 0 on the other
Find the roots of the corresponding quadratic equation
Use the sign of the leading coefficient to determine if the parabola is U-shaped ( $a > 0$ ) or n-shaped ( $a < 0$ )
Read the solution from the graph: $> 0$ where the graph is above the x-axis, $< 0$ where it is below

📐 Worked Example

Solve the inequality $2 x^{2} - x - 3 > 0$

1
The inequality is already rearranged. Factorise to find roots:
$2x^2 -x -3 = (2x -3)(x +1) = 0 \implies x = \frac{3}{2}, \quad x = -1$
2
This occurs outside the two roots for a U-shaped parabola
3
Write the final solution:
$x < -1 \quad \text{or} \quad x > \frac{3}{2}$

5. Simultaneous Equations: One Linear, One Quadratic★★★☆☆⏱ 4 min

When two equations must hold at once and one is linear while the other is quadratic, the reliable method is substitution: rearrange the linear equation to make one variable the subject, then substitute it into the quadratic. This collapses the pair into a single quadratic in one variable — which you already know how to solve.

📐 Worked Example

Solve the simultaneous equations $x + y + 1 = 0$ and $x^{2} + y^{2} = 25$ .

1
Make $y$ the subject of the linear equation:
$y = -x - 1$
2
Substitute into the quadratic equation:
$x^2 + (-x-1)^2 = 25$
3
Expand and collect into a standard quadratic:
$2x^2 + 2x - 24 = 0 \implies x^2 + x - 12 = 0$
4
Factorise and solve for $x$ :
$(x+4)(x-3) = 0 \implies x = -4 \ \text{or}\ x = 3$
5
Back-substitute each $x$ into the linear equation to pair the values:
$x=-4 \Rightarrow y=3; \quad x=3 \Rightarrow y=-4$

6. Disguised Quadratics: Equations Quadratic in a Function of x★★★★☆⏱ 4 min

Some equations are not quadratic in $x$ , yet they are quadratic in some function of $x$ — such as $x^{2}$ , $x$ , or $tan x$ . Substituting a single letter for that function reveals a standard quadratic. Solve for the substitute, then convert back to $x$ .

📐 Worked Example

Solve $x^{4} - 5 x^{2} + 4 = 0$ .

1
The equation is quadratic in $x^{2}$ . Let $u = x^{2}$ :
$u^2 - 5u + 4 = 0$
2
Solve the quadratic in $u$ :
$(u-1)(u-4) = 0 \implies u = 1 \ \text{or}\ u = 4$
3
Convert back with $x^{2} = u$ — each value can give two $x$ :
$x^2 = 1 \Rightarrow x = \pm 1; \quad x^2 = 4 \Rightarrow x = \pm 2$
4
State all four solutions:
$x = -2,\ -1,\ 1,\ 2$

7. Common Pitfalls

Wrong move:

Calling an expression quadratic when the coefficient of $x^{2}$ is zero

Why:

If $a = 0$ , the $x^{2}$ term disappears, leaving a linear expression, not a quadratic

Correct move:

Always confirm the coefficient of $x^{2}$ is non-zero before applying quadratic rules

Wrong move:

Mixing up the sign of $b$ when substituting into the quadratic formula

Why:

The formula starts with $- b$ , so a negative $b$ becomes positive after applying the negative sign

Correct move:

Write down $a, b, c$ with their explicit signs before substituting into the formula

Wrong move:

Stating that $Δ = 0$ means there are no real roots

Why:

A repeated root is still a real root, it just has two equal values

Correct move:

$Δ = 0$ means one repeated (equal) real root; $Δ < 0$ means no real roots

Wrong move:

For a U-shaped parabola, writing $a < x < b$ for $(x - p) (x - q) > 0$ where $p < q$

Why:

This is the solution for $< 0$ , the region between the roots for a U-shaped parabola

Correct move:

For $> 0$ (above the x-axis) on a U-shaped parabola, the solution is $x < p$ or $x > q$

8. Quick Reference Cheatsheet

Concept	Key Result	Use Case
General Form	$a x^{2} + b x + c, a \neq = 0$	Calculate discriminant, substitute into formula
Factorised Form	$a (x - p) (x - q)$	Identify roots, solve equations
Completed Square	$a (x - h)^{2} + k$	Find turning point at $(h, k)$
Quadratic Formula	$x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$	Solve any quadratic equation
2 distinct real roots	$Δ > 0$	Prove two distinct roots exist
Repeated real root	$Δ = 0$	Prove a line is tangent to a curve
No real roots	$Δ < 0$	Prove no intersection with x-axis
Quadratic Inequality ( $a > 0, p < q$ )	$(x - p) (x - q) < 0 ⟹ p < x < q$ ; $> 0 ⟹ x < p or x > q$	Quickly find solution ranges
Linear + quadratic simultaneous	Make a variable the subject of the linear equation, then substitute into the quadratic	Line-and-curve intersections, points of contact
Disguised quadratic	Let $u =$ a function of $x$ ; solve $a u^{2} + b u + c = 0$ , then convert back to $x$	$x^{4} - 5 x^{2} + 4 = 0$ , $6 x + x - 1 = 0$

Going deeper

syllabusCIE 9709 AS Pure 1 Syllabus

What's Next

Quadratics are the foundation of almost all other topics in CIE AS Pure 1, including curve sketching, coordinate geometry, and functions. The discriminant concept is regularly extended to problems asking for the number of intersections between a line and a curve, a common 5-6 mark question in Paper 1. Mastering quadratics helps you avoid losing easy method marks in more complex problems later in the course. The completed square form you learned here is also used extensively when finding the range of functions and sketching quadratic graphs.