# Quadratics

> CIE A-Level Mathematics · 9709 Pure Mathematics 1 (AS Level)
> Source: https://www.owlsprep.com/study/cie-9709-u1-quadratics/

This module covers all core concepts of quadratic expressions, equations, and inequalities for CIE 9709 AS Level Pure 1. You will learn to convert between forms, solve quadratics, and analyse root nature.

**Prerequisites:** Basic algebraic expansion and factorisation of linear terms; Solving single-variable linear equations

## Learning objectives

- Identify quadratic expressions and convert between different algebraic forms
- Solve quadratic equations using all standard methods
- Use the discriminant to determine the nature of roots of a quadratic
- Solve quadratic inequalities and state correct solution ranges
- Solve a pair of simultaneous equations where one is linear and one is quadratic, using substitution
- Recognise and solve equations that are quadratic in some function of x, such as x⁴ − 5x² + 4 = 0

## How often is this tested?

Based on OwlsPrep's analysis of official CIE 9709 P1 past papers (2016–2025): **Quadratics** appears **109 times** in the last 10 years — **14.9%** of all P1 questions (109 of 692).

Most-tested forms: Discriminant and nature of roots (50), Completing the square (39), Quadratic equations and graphs (19), Simultaneous equations (one linear, one quadratic) (17), Quadratic inequalities (6).

## Forms of Quadratic Expressions

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

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

*Example:* $3x^2 - 5x + 2$ is quadratic; $4x + 7$ is linear, so not quadratic.

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

- General form: $ax^2 + bx + c$, 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 $2x^2 - 8x + 3$ 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 $(h,k)$:

   $$(2, -5)$$

## Solving Quadratic Equations

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

**Comparing methods**

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

- **Factorisation** — Rearrange to $ax^2 +bx +c = 0$, 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 $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
  - 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 $2x^2 - 5x - 3 = 0$ 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:

   $$x = 3 \quad \text{or} \quad x = -\frac{1}{2}$$

## Discriminant and Nature of Roots

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

*Notation:* \Delta = b^2 - 4ac

*Example:* For $x^2 - 3x + 2$, $\Delta = 9 - 8 = 1$

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

> **tip**
>
> In CIE exams, any question asking you to 'show the equation has two distinct real roots' will always require you to calculate $\Delta$ and prove it is positive.

**Worked example:** Find the range of $k$ for which $x^2 + 2kx + 9 = 0$ has two distinct real roots

1. For two distinct real roots, we need $\Delta > 0$:

   $$\Delta = (2k)^2 - 4(1)(9) = 4k^2 - 36$$
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 $y = x^2 + 5x + 2$

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 $\Delta = 0$:

   $$\Delta = 4^2 - 4(1)(2 - k) = 0$$
3. Expand and solve for $k$:

   $$16 - 8 + 4k = 0 \implies 4k = -8 \implies k = -2$$

## Solving Quadratic Inequalities

To solve a quadratic inequality, follow this structured method:

1. Rearrange the inequality so all terms are on one side, with 0 on the other
2. Find the roots of the corresponding quadratic equation
3. Use the sign of the leading coefficient to determine if the parabola is U-shaped ($a>0$) or n-shaped ($a<0$)
4. 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 $2x^2 - x - 3 > 0$

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:

   $$x < -1 \quad \text{or} \quad x > \frac{3}{2}$$

## Simultaneous Equations: One Linear, One Quadratic

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:

   $$x=-4 \Rightarrow y=3; \quad x=3 \Rightarrow y=-4$$

> **tip**
>
> Substitute back into the linear equation, not the quadratic — it is simpler and avoids introducing extra solutions. Always pair each $x$ with the $y$ it produced: the solutions are the points $(-4, 3)$ and $(3, -4)$, not any other mix of these numbers.

## Disguised Quadratics: Equations Quadratic in a Function of x

Some equations are not quadratic in $x$, yet they are quadratic in some function of $x$ — such as $x^2$, $\sqrt{x}$, 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 $x^4 - 5x^2 + 4 = 0$.

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:

   $$x = -2,\ -1,\ 1,\ 2$$

> **warning**
>
> Solving for the substitute is only half the work — you must convert back to $x$, and each value may give 0, 1 or 2 values of $x$. Also watch the range of the substitution: in $6x + \sqrt{x} - 1 = 0$, put $u = \sqrt{x}$ so $6u^2 + u - 1 = 0$, giving $u = \tfrac{1}{3}$ or $u = -\tfrac{1}{2}$. Since $\sqrt{x} \ge 0$, reject $u = -\tfrac{1}{2}$, leaving $\sqrt{x} = \tfrac{1}{3}$, so $x = \tfrac{1}{9}$.

## Common pitfalls

- **Wrong:** Calling an expression quadratic when the coefficient of $x^2$ is zero
  - Why it fails: If $a=0$, the $x^2$ term disappears, leaving a linear expression, not a quadratic
  - Correct: Always confirm the coefficient of $x^2$ is non-zero before applying quadratic rules
- **Wrong:** Mixing up the sign of $b$ when substituting into the quadratic formula
  - Why it fails: The formula starts with $-b$, so a negative $b$ becomes positive after applying the negative sign
  - Correct: Write down $a, b, c$ with their explicit signs before substituting into the formula
- **Wrong:** Stating that $\Delta = 0$ means there are no real roots
  - Why it fails: A repeated root is still a real root, it just has two equal values
  - Correct: $\Delta = 0$ means one repeated (equal) real root; $\Delta < 0$ means no real roots
- **Wrong:** For a U-shaped parabola, writing $a < x < b$ for $(x-p)(x-q) > 0$ where $p<q$
  - Why it fails: This is the solution for $<0$, the region between the roots for a U-shaped parabola
  - Correct: For $>0$ (above the x-axis) on a U-shaped parabola, the solution is $x < p$ or $x > q$

## Cheatsheet

| Concept | Key Result | Use Case |
| --- | --- | --- |
| General Form | $ax^2 + bx + c, a \neq 0$ | 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 | $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ | Solve any quadratic equation |
| 2 distinct real roots | $\Delta > 0$ | Prove two distinct roots exist |
| Repeated real root | $\Delta = 0$ | Prove a line is tangent to a curve |
| No real roots | $\Delta < 0$ | Prove no intersection with x-axis |
| Quadratic Inequality ($a>0, p<q$) | $(x-p)(x-q) < 0 \implies p<x<q$; $>0 \implies x<p \text{ or } x>q$ | 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 $au^2+bu+c=0$, then convert back to $x$ | $x^4-5x^2+4=0$, $\;6x+\sqrt{x}-1=0$ |

## 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.

- [Functions](https://www.owlsprep.com/study/cie-9709-u1-functions/)
- [Coordinate Geometry](https://www.owlsprep.com/study/cie-9709-u1-coordinate-geometry/)
- [Circular measure](https://www.owlsprep.com/study/cie-9709-u1-circular-measure/)

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