AlgebraAQA GCSE Mathematics (8300)

Solving Equations and Inequalities

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

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Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Solving a quadratic by completing the square
Higher
Question
Solve $x^2 - 8x + 11 = 0$ , giving your answers in surd form.
Solution
1. 1
  Write the left-hand side in completed-square form.
  $(x - 4)^2 - 16 + 11 = 0$
2. 2
  Simplify the constant terms.
  $(x - 4)^2 - 5 = 0$
3. 3
  Isolate the squared bracket.
  $(x - 4)^2 = 5$
4. 4
  Take the square root of both sides (±).
  $x - 4 = \pm\sqrt{5}$
5. 5
  Add 4 to both sides for the final answer.
  $x = 4 \pm \sqrt{5}$
Answer
$x = 4 + \sqrt{5}$ or $x = 4 - \sqrt{5}$
02.Solving simultaneous equations (linear and quadratic)
Challenge
Question
Solve simultaneously: $y = 2x - 3$ and $x^2 + y^2 = 13$ .
Solution
1. 1
  Substitute the linear expression for $y$ into the circle equation.
  $x^2 + (2x - 3)^2 = 13$
2. 2
  Expand the bracket.
  $x^2 + 4x^2 - 12x + 9 = 13$
3. 3
  Collect all terms on one side and simplify.
  $5x^2 - 12x - 4 = 0$
4. 4
  Factorise.
  $(5x + 2)(x - 2) = 0$
5. 5
  Solve for $x$ and find corresponding $y$ values using $y = 2x - 3$ .
  $x = -\tfrac{2}{5},\; y = -\tfrac{19}{5} \quad \text{or} \quad x = 2,\; y = 1$
Answer
$x = -\dfrac{2}{5},\; y = -\dfrac{19}{5}$ or $x = 2,\; y = 1$
03.Solving a quadratic inequality
Higher
Question
Find the set of values of $x$ for which $x^2 - 3x - 10 > 0$ .
Solution
1. 1
  Solve the corresponding equation to find the critical values.
  $x^2 - 3x - 10 = 0 \implies (x - 5)(x + 2) = 0$
2. 2
  The critical values are $x = 5$ and $x = -2$ .
3. 3
  The coefficient of $x^2$ is positive, so the parabola opens upward. The expression is greater than zero outside the roots.
4. 4
  Write the solution set.
  $x < -2 \text{ or } x > 5$
Answer
$\{x : x < -2\} \cup \{x : x > 5\}$
04.Iteration to find a root to 3 decimal places
Challenge
Question
The equation $x^3 + x = 7$ has one real root. Show it lies between 1 and 2, then use $x_{n+1} = \sqrt[3]{7 - x_n}$ with $x_0 = 1.6$ to find the root to 3 decimal places.
Solution
1. 1
  Let $f(x) = x^3 + x - 7$ . Evaluate at the endpoints.
  $f(1) = 1 + 1 - 7 = -5 < 0; \quad f(2) = 8 + 2 - 7 = 3 > 0$
2. 2
  There is a change of sign, so a root lies in $(1, 2)$ .
3. 3
  Apply the iterative formula from $x_0 = 1.6$ .
  $x_1 = \sqrt[3]{7 - 1.6} = \sqrt[3]{5.4} \approx 1.7520$
4. 4
  Continue iterating.
  $x_2 \approx \sqrt[3]{7 - 1.7520} \approx 1.7218; \quad x_3 \approx 1.7282; \quad x_4 \approx 1.7269; \quad x_5 \approx 1.7272$
5. 5
  The sequence converges. Consecutive values both round to 1.727 to 3 d.p.
  $x \approx 1.727$
Answer
Root $\approx 1.727$ (3 d.p.)

Key formulae

Quadratic Formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
- $a$
  — coefficient of $x^2$
- $b$
  — coefficient of $x$
- $c$
  — constant term
When to use
Solving any quadratic $ax^2 + bx + c = 0$ ; particularly when factorising is not straightforward or when exact surd answers are required.
Discriminant
$\Delta = b^2 - 4ac$
- $\Delta > 0$
  — two distinct real roots
- $\Delta = 0$
  — one repeated real root
- $\Delta < 0$
  — no real roots
When to use
Determining the number of real solutions without fully solving the equation; used in 'show that the equation has no real solutions' questions.
Completed Square Form
$ax^2 + bx + c = a\!\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$
- $\left(x + \frac{b}{2a}\right)^2$
  — squared bracket
- $-\frac{b^2-4ac}{4a}$
  — adjustment constant
When to use
Solving quadratics in exact form, finding the vertex of a parabola, or proving results about the minimum/maximum of a quadratic.
Iterative Sequence
$x_{n+1} = f(x_n)$
- $x_0$
  — initial estimate (starting value)
- $x_{n+1}$
  — next approximation
- $f(x_n)$
  — rearrangement of the equation
When to use
Finding approximate roots of equations numerically when no exact algebraic method is practical; the formula is always given in the AQA question.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Root (of an equation): A value of the variable that satisfies the equation, i.e. makes both sides equal. Graphically, a root of $f(x) = 0$ is the $x$ -coordinate where the graph crosses the $x$ -axis.
Discriminant: The expression $b^2 - 4ac$ for a quadratic $ax^2 + bx + c$ . Its sign determines whether the quadratic has two real roots, one repeated root, or no real roots.
Simultaneous equations: A set of two or more equations that share the same unknown variables and must all be satisfied at the same time. The solution gives the value(s) of each variable.
Iteration: A process of repeatedly applying the same operation to successive results. In the context of solving equations, an iterative formula $x_{n+1} = f(x_n)$ generates a sequence that converges to a root.
Inequality: A mathematical statement showing that two expressions are not necessarily equal, using the symbols $<$ , $>$ , $\leq$ or $\geq$ . The solution is a range of values, not a single value.
Set notation: A formal way to describe solution sets: $\{x : x > 3\}$ means 'the set of all real numbers $x$ such that $x > 3$ '. The union symbol $\cup$ is used when the solution consists of two separate intervals.

Common mistakes and misconceptions

Mistake
Forgetting to reverse the inequality sign when dividing by a negative number
Why it happens
Students apply the same rule as solving equations (do the same to both sides) without remembering that multiplying or dividing by a negative reverses the order of the number line.
How to avoid it
Write a reminder next to any step where you divide/multiply by a negative, and check by substituting a value from your answer into the original inequality.
Mistake
Applying the quadratic formula incorrectly — errors with the $\pm$ and the $2a$ denominator
Why it happens
Students write $-b \pm \sqrt{b^2 - 4ac} / 2a$ without brackets, so only the square root (not the whole numerator) is divided by $2a$ .
How to avoid it
Write the formula with explicit brackets: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Substitute values before simplifying, and check both roots satisfy the original equation.
Mistake
Writing the wrong region for a quadratic inequality
Why it happens
Students find the roots correctly but then choose the wrong region — e.g. writing $x < 2$ and $x > 5$ for an inequality that should give $2 < x < 5$ .
How to avoid it
Always sketch the parabola. For an upward-opening parabola: $< 0$ means between the roots; $> 0$ means outside the roots. Test a value in each region to confirm.
Mistake
Not pairing simultaneous solutions correctly when both equations give two $x$ values
Why it happens
After solving a linear-quadratic system, students list $x_1, x_2$ and $y_1, y_2$ separately without pairing the correct $x$ with its corresponding $y$ .
How to avoid it
For each $x$ value found from the quadratic, immediately substitute back into the linear equation to find the paired $y$ value. Present solutions as coordinate pairs or clearly labelled solution sets.

AlgebraAQA GCSE Mathematics (8300)

Solving Equations and Inequalities

Work through the notes, try the practice questions, then take the quiz. The report tells you exactly what to revise next. (2026)

Previous Next

Master the topic

Worked examples4 Key formulae4 Definitions6 Common mistakes4

Step-by-step worked examples

01.Solving a quadratic by completing the square
Higher
Question
Solve $x^2 - 8x + 11 = 0$ , giving your answers in surd form.
Solution
1. 1
  Write the left-hand side in completed-square form.
  $(x - 4)^2 - 16 + 11 = 0$
2. 2
  Simplify the constant terms.
  $(x - 4)^2 - 5 = 0$
3. 3
  Isolate the squared bracket.
  $(x - 4)^2 = 5$
4. 4
  Take the square root of both sides (±).
  $x - 4 = \pm\sqrt{5}$
5. 5
  Add 4 to both sides for the final answer.
  $x = 4 \pm \sqrt{5}$
Answer
$x = 4 + \sqrt{5}$ or $x = 4 - \sqrt{5}$
02.Solving simultaneous equations (linear and quadratic)
Challenge
Question
Solve simultaneously: $y = 2x - 3$ and $x^2 + y^2 = 13$ .
Solution
1. 1
  Substitute the linear expression for $y$ into the circle equation.
  $x^2 + (2x - 3)^2 = 13$
2. 2
  Expand the bracket.
  $x^2 + 4x^2 - 12x + 9 = 13$
3. 3
  Collect all terms on one side and simplify.
  $5x^2 - 12x - 4 = 0$
4. 4
  Factorise.
  $(5x + 2)(x - 2) = 0$
5. 5
  Solve for $x$ and find corresponding $y$ values using $y = 2x - 3$ .
  $x = -\tfrac{2}{5},\; y = -\tfrac{19}{5} \quad \text{or} \quad x = 2,\; y = 1$
Answer
$x = -\dfrac{2}{5},\; y = -\dfrac{19}{5}$ or $x = 2,\; y = 1$
03.Solving a quadratic inequality
Higher
Question
Find the set of values of $x$ for which $x^2 - 3x - 10 > 0$ .
Solution
1. 1
  Solve the corresponding equation to find the critical values.
  $x^2 - 3x - 10 = 0 \implies (x - 5)(x + 2) = 0$
2. 2
  The critical values are $x = 5$ and $x = -2$ .
3. 3
  The coefficient of $x^2$ is positive, so the parabola opens upward. The expression is greater than zero outside the roots.
4. 4
  Write the solution set.
  $x < -2 \text{ or } x > 5$
Answer
$\{x : x < -2\} \cup \{x : x > 5\}$
04.Iteration to find a root to 3 decimal places
Challenge
Question
The equation $x^3 + x = 7$ has one real root. Show it lies between 1 and 2, then use $x_{n+1} = \sqrt[3]{7 - x_n}$ with $x_0 = 1.6$ to find the root to 3 decimal places.
Solution
1. 1
  Let $f(x) = x^3 + x - 7$ . Evaluate at the endpoints.
  $f(1) = 1 + 1 - 7 = -5 < 0; \quad f(2) = 8 + 2 - 7 = 3 > 0$
2. 2
  There is a change of sign, so a root lies in $(1, 2)$ .
3. 3
  Apply the iterative formula from $x_0 = 1.6$ .
  $x_1 = \sqrt[3]{7 - 1.6} = \sqrt[3]{5.4} \approx 1.7520$
4. 4
  Continue iterating.
  $x_2 \approx \sqrt[3]{7 - 1.7520} \approx 1.7218; \quad x_3 \approx 1.7282; \quad x_4 \approx 1.7269; \quad x_5 \approx 1.7272$
5. 5
  The sequence converges. Consecutive values both round to 1.727 to 3 d.p.
  $x \approx 1.727$
Answer
Root $\approx 1.727$ (3 d.p.)

Key formulae

Quadratic Formula
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
- $a$
  — coefficient of $x^2$
- $b$
  — coefficient of $x$
- $c$
  — constant term
When to use
Solving any quadratic $ax^2 + bx + c = 0$ ; particularly when factorising is not straightforward or when exact surd answers are required.
Discriminant
$\Delta = b^2 - 4ac$
- $\Delta > 0$
  — two distinct real roots
- $\Delta = 0$
  — one repeated real root
- $\Delta < 0$
  — no real roots
When to use
Determining the number of real solutions without fully solving the equation; used in 'show that the equation has no real solutions' questions.
Completed Square Form
$ax^2 + bx + c = a\!\left(x + \frac{b}{2a}\right)^2 - \frac{b^2 - 4ac}{4a}$
- $\left(x + \frac{b}{2a}\right)^2$
  — squared bracket
- $-\frac{b^2-4ac}{4a}$
  — adjustment constant
When to use
Solving quadratics in exact form, finding the vertex of a parabola, or proving results about the minimum/maximum of a quadratic.
Iterative Sequence
$x_{n+1} = f(x_n)$
- $x_0$
  — initial estimate (starting value)
- $x_{n+1}$
  — next approximation
- $f(x_n)$
  — rearrangement of the equation
When to use
Finding approximate roots of equations numerically when no exact algebraic method is practical; the formula is always given in the AQA question.

Practice with flashcards

Card 1 of 60 known · 0 to review

Key definitions and keywords

Root (of an equation): A value of the variable that satisfies the equation, i.e. makes both sides equal. Graphically, a root of $f(x) = 0$ is the $x$ -coordinate where the graph crosses the $x$ -axis.
Discriminant: The expression $b^2 - 4ac$ for a quadratic $ax^2 + bx + c$ . Its sign determines whether the quadratic has two real roots, one repeated root, or no real roots.
Simultaneous equations: A set of two or more equations that share the same unknown variables and must all be satisfied at the same time. The solution gives the value(s) of each variable.
Iteration: A process of repeatedly applying the same operation to successive results. In the context of solving equations, an iterative formula $x_{n+1} = f(x_n)$ generates a sequence that converges to a root.
Inequality: A mathematical statement showing that two expressions are not necessarily equal, using the symbols $<$ , $>$ , $\leq$ or $\geq$ . The solution is a range of values, not a single value.
Set notation: A formal way to describe solution sets: $\{x : x > 3\}$ means 'the set of all real numbers $x$ such that $x > 3$ '. The union symbol $\cup$ is used when the solution consists of two separate intervals.

Common mistakes and misconceptions

Mistake
Forgetting to reverse the inequality sign when dividing by a negative number
Why it happens
Students apply the same rule as solving equations (do the same to both sides) without remembering that multiplying or dividing by a negative reverses the order of the number line.
How to avoid it
Write a reminder next to any step where you divide/multiply by a negative, and check by substituting a value from your answer into the original inequality.
Mistake
Applying the quadratic formula incorrectly — errors with the $\pm$ and the $2a$ denominator
Why it happens
Students write $-b \pm \sqrt{b^2 - 4ac} / 2a$ without brackets, so only the square root (not the whole numerator) is divided by $2a$ .
How to avoid it
Write the formula with explicit brackets: $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ . Substitute values before simplifying, and check both roots satisfy the original equation.
Mistake
Writing the wrong region for a quadratic inequality
Why it happens
Students find the roots correctly but then choose the wrong region — e.g. writing $x < 2$ and $x > 5$ for an inequality that should give $2 < x < 5$ .
How to avoid it
Always sketch the parabola. For an upward-opening parabola: $< 0$ means between the roots; $> 0$ means outside the roots. Test a value in each region to confirm.
Mistake
Not pairing simultaneous solutions correctly when both equations give two $x$ values
Why it happens
After solving a linear-quadratic system, students list $x_1, x_2$ and $y_1, y_2$ separately without pairing the correct $x$ with its corresponding $y$ .
How to avoid it
For each $x$ value found from the quadratic, immediately substitute back into the linear equation to find the paired $y$ value. Present solutions as coordinate pairs or clearly labelled solution sets.

Solving Equations and Inequalities

Master the topic

Step-by-step worked examples

01.Solving a quadratic by completing the square

02.Solving simultaneous equations (linear and quadratic)

03.Solving a quadratic inequality

04.Iteration to find a root to 3 decimal places

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions

Solving Equations and Inequalities

Master the topic

Step-by-step worked examples

01.Solving a quadratic by completing the square

02.Solving simultaneous equations (linear and quadratic)

03.Solving a quadratic inequality

04.Iteration to find a root to 3 decimal places

Key formulae

Practice with flashcards

Key definitions and keywords

Common mistakes and misconceptions