Solving Quadratic Equations and Inequalities (9709 Pure Mathematics 1)

Introduction

Quadratic equations and inequalities are among the most frequently tested topics across all 9709 papers. Mastery here is non-negotiable: they arise independently as well as embedded within harder problems involving functions, coordinate geometry, and calculus. The syllabus requires you to solve $ax^2 + bx + c = 0$ (and the corresponding inequality) using three distinct methods: factorising, completing the square, and the quadratic formula. Examiners expect you to choose the most efficient method and to present solutions with full algebraic justification.

Core Concept

A quadratic equation in one unknown $x$ has the form

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

and has at most two real solutions (roots). A quadratic inequality asks for the set of values of $x$ for which $ax^2 + bx + c > 0$ , $\geq 0$ , $< 0$ , or $\leq 0$ .

Method 1 — Factorising

Express the left-hand side as a product of two linear factors:

ax^2 + bx + c = a(x - \alpha)(x - \beta)

Setting each factor to zero gives the roots $x = \alpha$ and $x = \beta$ . This method is fastest when integer or simple fractional roots exist.

Method 2 — Completing the Square

Rewrite the equation in the form $a(x+p)^2 + q = 0$ , then isolate $x$ :

x + p = \pm\sqrt{-\tfrac{q}{a}} \implies x = -p \pm \sqrt{-\tfrac{q}{a}}

This method is essential when the equation does not factorise neatly and is the basis for understanding the vertex form of a quadratic (a prerequisite skill you should already have).

Method 3 — The Quadratic Formula

Always applicable. For $ax^2 + bx + c = 0$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The discriminant $\Delta = b^2 - 4ac$ determines the nature of the roots (a prerequisite topic). If $\Delta < 0$ there are no real solutions.

Solving Quadratic Inequalities

To solve $ax^2 + bx + c < 0$ (or $> 0$ ):

Find the roots $\alpha \leq \beta$ by solving $ax^2 + bx + c = 0$ .
Sketch (or reason from) the parabola's shape ( $a > 0$ opens upward; $a < 0$ opens downward).
Read off the solution set:
- $a > 0$ : the parabola is below the axis between the roots, so $ax^2+bx+c < 0 \Rightarrow \alpha < x < \beta$ .
- $a > 0$ : the parabola is above the axis outside the roots, so $ax^2+bx+c > 0 \Rightarrow x < \alpha \text{ or } x > \beta$ .
- Reverse the regions when $a < 0$ .

Key Formulae & Definitions

\boxed{ax^2 + bx + c = 0 \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}}

\text{Completed-square form: } a\!\left(x + \frac{b}{2a}\right)^{\!2} + c - \frac{b^2}{4a} = 0

Discriminant $\Delta = b^2 - 4ac$	Nature of roots
$\Delta > 0$	Two distinct real roots
$\Delta = 0$	One repeated real root
$\Delta < 0$	No real roots

Solution sets for inequalities (assuming roots $\alpha < \beta$ , $a > 0$ ):

ax^2+bx+c < 0 \iff \alpha < x < \beta

ax^2+bx+c > 0 \iff x < \alpha \;\text{ or }\; x > \beta

Worked Examples

Example 1 — Solving by Factorising

Solve $2x^2 - 7x + 3 = 0$ .

Step 1. Identify $a=2,\, b=-7,\, c=3$ . We need two numbers that multiply to $ac = 6$ and add to $b = -7$ : these are $-6$ and $-1$ .

Step 2. Split the middle term:

2x^2 - 6x - x + 3 = 0

Step 3. Factor by grouping:

2x(x - 3) - 1(x - 3) = 0 \implies (2x - 1)(x - 3) = 0

Step 4. Set each factor to zero:

2x - 1 = 0 \implies x = \tfrac{1}{2}, \qquad x - 3 = 0 \implies x = 3

\boxed{x = \tfrac{1}{2} \text{ or } x = 3}

Example 2 — Solving by Completing the Square

Solve $3x^2 - 6x - 2 = 0$ , giving answers in surd form.

Step 1. Divide through by 3:

x^2 - 2x - \tfrac{2}{3} = 0

Step 2. Complete the square on $x^2 - 2x$ :

\left(x - 1\right)^2 - 1 - \tfrac{2}{3} = 0 \implies (x-1)^2 = \tfrac{5}{3}

Step 3. Take the square root of both sides:

x - 1 = \pm\sqrt{\tfrac{5}{3}} = \pm\frac{\sqrt{15}}{3}

Step 4. Solve for $x$ :

\boxed{x = 1 \pm \frac{\sqrt{15}}{3}}

Example 3 — Solving a Quadratic Inequality

Solve $x^2 - x - 6 < 0$ .

Step 1. Solve the equation $x^2 - x - 6 = 0$ :

(x-3)(x+2) = 0 \implies x = 3 \text{ or } x = -2

Step 2. Since $a = 1 > 0$ the parabola opens upward, so the quadratic is negative between the roots.

Step 3. Write the solution set:

\boxed{-2 < x < 3}

Example 4 — Using the Formula, then an Inequality

Find the set of values of $x$ for which $2x^2 + 3x - 4 > 0$ .

Step 1. Apply the quadratic formula to $2x^2 + 3x - 4 = 0$ :

x = \frac{-3 \pm \sqrt{9 + 32}}{4} = \frac{-3 \pm \sqrt{41}}{4}

So $\alpha = \dfrac{-3 - \sqrt{41}}{4} \approx -2.35$ and $\beta = \dfrac{-3+\sqrt{41}}{4} \approx 0.85$ .

Step 2. Since $a = 2 > 0$ the parabola opens upward, so the quadratic is positive outside the roots:

\boxed{x < \frac{-3-\sqrt{41}}{4} \quad \text{or} \quad x > \frac{-3+\sqrt{41}}{4}}

Common Mistakes & Examiner Pitfalls

Dividing by $x$ when solving equations. Writing $ax^2 + bx = 0$ as $ax + b = 0$ by dividing through by $x$ loses the root $x = 0$ . Always factorise: $x(ax + b) = 0$ .
Wrong inequality direction when multiplying/dividing by a negative. This applies when rearranging, not when using the parabola method. Use the parabola sketch approach to avoid sign errors entirely.
Using " $\text{or}$ " instead of " $\text{and}$ " for the inner region. The solution to $x^2 < 9$ is $-3 < x < 3$ (a single compound inequality), not $x < 3$ or $x > -3$ .
Forgetting $\pm$ when square-rooting. In completing the square, both the positive and negative square root must be taken.
Arithmetic slips in the discriminant. With $b$ negative, $b^2$ is always positive — take care with signs when computing $b^2 - 4ac$ under exam pressure.
Leaving the answer as two separate equations for an inequality. The final answer must be a set of values (interval notation or inequality notation), not " $x = \ldots$ ".

Practice Questions

Q1. Solve $6x^2 + x - 12 = 0$ by factorising.

<details><summary>Show answer</summary>

Find two numbers multiplying to $6 \times (-12) = -72$ and adding to $1$ : these are $9$ and $-8$ .

6x^2 + 9x - 8x - 12 = 0

3x(2x+3) - 4(2x+3) = 0

(3x-4)(2x+3) = 0

x = \tfrac{4}{3} \quad \text{or} \quad x = -\tfrac{3}{2}

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Q2. Solve $x^2 + 6x + 2 = 0$ by completing the square, leaving your answer in surd form.

<details><summary>Show answer</summary>

(x+3)^2 - 9 + 2 = 0

(x+3)^2 = 7

x + 3 = \pm\sqrt{7}

x = -3 \pm \sqrt{7}

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Q3. Use the quadratic formula to solve $3x^2 - 5x + 1 = 0$ , giving your answers correct to 3 significant figures.

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x = \frac{5 \pm \sqrt{25 - 12}}{6} = \frac{5 \pm \sqrt{13}}{6}

$\sqrt{13} \approx 3.606$

x = \frac{5 + 3.606}{6} \approx 1.43 \quad \text{or} \quad x = \frac{5-3.606}{6} \approx 0.232

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Q4. Find the set of values of $x$ satisfying $x^2 \leq 5x - 4$ .

<details><summary>Show answer</summary>

Rearrange: $x^2 - 5x + 4 \leq 0$

Factorise: $(x-1)(x-4) \leq 0$

Roots: $x = 1$ and $x = 4$ . Since $a = 1 > 0$ , the parabola opens upward, so the expression is non-positive between the roots (endpoints included as the inequality is $\leq$ ):

1 \leq x \leq 4

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Q5. Find the set of values of $x$ for which $2x^2 - 7x + 3 > 0$ .

<details><summary>Show answer</summary>

From Example 1, the roots of $2x^2 - 7x + 3 = 0$ are $x = \tfrac{1}{2}$ and $x = 3$ .

Since $a = 2 > 0$ , the parabola opens upward, so $2x^2 - 7x + 3 > 0$ outside the roots:

x < \tfrac{1}{2} \quad \text{or} \quad x > 3

</details>

Connections

Prerequisites you should be confident with:

Completing the Square — the algebraic technique used directly in Method 2 and for deriving the quadratic formula.
The Discriminant — determines in advance how many real solutions exist, guiding your choice of method.

Topics this directly enables:

Quadratic Functions and Graphs — understanding roots as $x$ -intercepts of $y = ax^2 + bx + c$ .
Simultaneous Equations — solving a linear and quadratic pair leads to a quadratic equation.
Functions: Domain and Range — quadratic inequalities define restricted domains.
Differentiation Applications — finding intervals where a derivative is positive/negative uses the same inequality technique.

Introduction

Core Concept

Method 1 — Factorising

Method 2 — Completing the Square

Method 3 — The Quadratic Formula

Solving Quadratic Inequalities

Key Formulae & Definitions

Worked Examples

Example 1 — Solving by Factorising

Example 2 — Solving by Completing the Square

Example 3 — Solving a Quadratic Inequality

Example 4 — Using the Formula, then an Inequality

Common Mistakes & Examiner Pitfalls

Practice Questions

Connections

Figures

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