Solve: \(x^2 = 25\)
Remember that both positive and negative numbers can square to 25.
\(x^2 = 25\), so \(x = \pm 5\). Therefore, \(x = -5\) or \(x = 5\).
Quadratic Equations
Quadratic equations involve squared terms and can be solved using methods such as factorising. These equations are a key part of higher GCSE Maths.
Overview
A quadratic equation is an equation where the highest power of \(x\) is 2.
The general form is:
\( ax^2 + bx + c = 0 \)
Solving a quadratic means finding the value or values of \(x\) that make the equation true.
These values are called the roots or solutions.
What you should understand after this topic
- Understand what makes an equation quadratic
- Understand what roots or solutions mean
- Solve quadratic equations by factorising
- Solve quadratic equations using the quadratic formula
- Understand how graphs connect to solutions
Key Definitions
Quadratic Equation
An equation where the highest power of the variable is 2.
Root / Solution
A value of \(x\) that makes the quadratic equal to 0.
Factorising
Writing the quadratic as a product of two brackets.
Quadratic Formula
A formula that can be used to solve any quadratic equation.
Discriminant
The part \(b^2 - 4ac\), which tells you how many real solutions there are.
Parabola
The curved graph shape made by a quadratic function.
Key Rules
How to Solve
Step 1: Understand what solving means
Solving a quadratic means finding the values of \(x\) that make the equation equal to zero.
\( ax^2 + bx + c = 0 \)
On a graph, the solutions are the x-intercepts.
See quadratic graphs for the graph link.
Step 2: Identify \(a\), \(b\), and \(c\)
Write the quadratic in standard form before choosing a method.
\( ax^2 + bx + c = 0 \)
\(a\) = coefficient of \(x^2\)
\(b\) = coefficient of \(x\)
\(c\) = constant term
Step 3: Try factorising first
Factorising is usually the quickest method when the quadratic splits into brackets.
\( x^2 + 5x + 6 = (x+2)(x+3) \)
Then set each bracket equal to zero.
\(x+2=0\) or \(x+3=0\)
Answer: \(x=-2\) or \(x=-3\)
Step 4: Use the quadratic formula when needed
The quadratic formula works for any quadratic equation.
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
Exam tip: Use the formula when factorising is difficult or when decimals are expected.
Step 5: Use the discriminant
The discriminant tells you how many real solutions the quadratic has.
\( b^2 - 4ac \)
Positive
Two real solutions.
Zero
One repeated solution.
Negative
No real solutions.
Step 6: Link solutions to the graph
Exam thinking: Algebra gives exact roots; the graph shows where they occur.
Two solutions
Graph crosses the x-axis twice.
One solution
Graph touches the x-axis once.
No real solutions
Graph does not cross the x-axis.
Step 7: Exam method summary
- Put the equation into the form \(ax^2 + bx + c = 0\).
- Identify \(a\), \(b\), and \(c\).
- Try factorising if the numbers are simple.
- Use the quadratic formula if factorising is not suitable.
- Check whether the answer needs exact form or decimals.
Example Questions
Edexcel
Exam-style questions inspired by Edexcel GCSE Mathematics, focusing on solving quadratic equations using factorisation and interpretation.
Edexcel
Solve \( x^2 + 5x + 6 = 0 \).
Edexcel
Solve \( x^2 - 9 = 0 \).
Edexcel
Solve \( x^2 - 4x - 12 = 0 \).
Edexcel
Solve \( 2x^2 + 7x + 3 = 0 \).
Edexcel
The area of a rectangle is 24 cm². Its length is \( x+2 \) cm and its width is \( x \) cm. Form and solve a quadratic equation to find the value of \( x \).
AQA
Exam-style questions based on the AQA GCSE Mathematics specification, focusing on solving quadratics by completing the square and using the quadratic formula.
AQA
Solve \( x^2 + 6x + 5 = 0 \) by factorising.
AQA
Solve \( x^2 + 4x - 1 = 0 \) correct to 2 decimal places.
AQA
Solve \( x^2 + 8x + 3 = 0 \) using the quadratic formula.
AQA
Solve \( (x - 3)^2 = 16 \).
AQA
Show that the solutions of \( x^2 - 2x - 8 = 0 \) are \( x = 4 \) and \( x = -2 \).
OCR
Exam-style questions aligned with OCR GCSE Mathematics, emphasising reasoning, algebraic manipulation, and interpreting solutions.
OCR
Solve \( 3x^2 - 12x = 0 \).
OCR
Solve \( 2x^2 - 5x - 3 = 0 \).
OCR
Find the values of \( k \) for which \( x^2 + kx + 9 = 0 \) has equal roots.
OCR
The equation \( x^2 - px + 12 = 0 \) has roots 3 and 4. Find the value of \( p \).
OCR
Explain why the equation \( x^2 + 4x + 5 = 0 \) has no real solutions.
Exam Checklist
Step 1
Write the equation in the form \(ax^2 + bx + c = 0\).
Step 2
Choose the best method: factorise, formula or graph.
Step 3
Solve carefully and show your working clearly.
Step 4
Check both answers and watch signs carefully.
Most common exam mistakes
Factorising
Wrong pair of numbers or wrong signs in the brackets.
Formula
Substituting the wrong values for \(a\), \(b\), or \(c\).
Discriminant
Forgetting what the result tells you about the number of roots.
Answers
Forgetting that a quadratic often has two solutions.
Common Mistakes
These are common mistakes students make when solving quadratic equations in GCSE Maths.
Not writing in standard form
Incorrect
A student tries to solve without rearranging the equation.
Correct
Always rewrite the equation in the form \(ax^2 + bx + c = 0\) before solving.
Choosing the wrong factors
Incorrect
A student picks numbers that do not multiply and add correctly.
Correct
When factorising, choose two numbers that multiply to \(ac\) and add to \(b\). Check both conditions.
Forgetting both solutions
Incorrect
A student finds only one value of x.
Correct
Each bracket gives a solution. For example, \((x - 2)(x + 3) = 0\) gives \(x = 2\) and \(x = -3\).
Incorrect substitution into the formula
Incorrect
A student substitutes incorrect values for a, b or c.
Correct
Carefully identify a, b and c from the standard form before substituting into the quadratic formula.
Errors with negative numbers
Incorrect
A student makes sign mistakes during calculation.
Correct
Pay close attention to signs, especially when substituting into \(-b\) and when squaring \(b\).
Forgetting the ± sign
Incorrect
A student writes only one answer from the quadratic formula.
Correct
The \(\pm\) sign gives two solutions. Always calculate both the positive and negative cases.
Try It Yourself
Practise solving quadratic equations using different methods.
Foundation
Foundation Practice
Solve simple quadratic equations by factorising or using square roots.
Solve: \(x^2 = 49\). Give both answers separated by a comma.
Find both square roots of 49.
\(x^2 = 49\), so \(x = -7\) or \(x = 7\).
Solve: \(x^2 - 9 = 0\)
Rearrange to \(x^2 = 9\).
\(x^2 - 9 = 0\). Add 9 to both sides: \(x^2 = 9\). So \(x = -3\) or \(x = 3\).
Solve: \(x^2 - 16 = 0\). Give both answers separated by a comma.
Move 16 to the other side.
\(x^2 - 16 = 0\), so \(x^2 = 16\). Therefore, \(x = -4\) or \(x = 4\).
Solve: \((x + 2)(x + 5) = 0\)
If two brackets multiply to 0, one bracket must be 0.
Set each bracket equal to 0: \(x + 2 = 0\) gives \(x = -2\), and \(x + 5 = 0\) gives \(x = -5\).
Solve: \((x - 3)(x + 4) = 0\). Give both answers separated by a comma.
Set each bracket equal to zero.
\(x - 3 = 0\) gives \(x = 3\). \(x + 4 = 0\) gives \(x = -4\).
Solve: \(x^2 + 5x + 6 = 0\)
Factorise the quadratic first.
\(x^2 + 5x + 6 = (x + 2)(x + 3)\). So \(x = -2\) or \(x = -3\).
Solve: \(x^2 - 7x + 12 = 0\). Give both answers separated by a comma.
Find two numbers that multiply to 12 and add to -7.
\(x^2 - 7x + 12 = (x - 3)(x - 4)\). So \(x = 3\) or \(x = 4\).
A student solves \(x^2 = 36\) and writes only \(x = 6\). What did they forget?
Both \(6^2\) and \((-6)^2\) equal 36.
\(x^2 = 36\) has two solutions: \(x = -6\) and \(x = 6\).
Solve: \(x^2 + 2x - 8 = 0\). Give both answers separated by a comma.
Find two numbers that multiply to -8 and add to 2.
\(x^2 + 2x - 8 = (x + 4)(x - 2)\). So \(x = -4\) or \(x = 2\).
Higher
Higher Practice
Solve quadratics by factorising, rearranging and using the quadratic formula.
Solve: \(2x^2 + 7x + 3 = 0\)
Factorise into two brackets.
\(2x^2 + 7x + 3 = (2x + 1)(x + 3)\). So \(2x + 1 = 0\) gives \(x = -\frac{1}{2}\), and \(x + 3 = 0\) gives \(x = -3\).
Solve: \(3x^2 + 10x + 8 = 0\). Give both answers separated by a comma.
Factorise into \((3x + 4)(x + 2)\).
\(3x^2 + 10x + 8 = (3x + 4)(x + 2)\). So \(x = -\frac{4}{3}\) or \(x = -2\).
Solve: \(x^2 - 5x = 0\)
Factorise by taking out x.
\(x^2 - 5x = x(x - 5)\). So \(x = 0\) or \(x = 5\).
Solve: \(2x^2 - 8x = 0\). Give both answers separated by a comma.
Factorise by taking out \(2x\).
\(2x^2 - 8x = 2x(x - 4)\). So \(x = 0\) or \(x = 4\).
Solve: \(x^2 - 4x - 12 = 0\)
Find two numbers that multiply to -12 and add to -4.
\(x^2 - 4x - 12 = (x - 6)(x + 2)\). So \(x = 6\) or \(x = -2\).
Solve: \(x^2 + 6x + 5 = 0\). Give both answers separated by a comma.
Find two numbers that multiply to 5 and add to 6.
\(x^2 + 6x + 5 = (x + 1)(x + 5)\). So \(x = -1\) or \(x = -5\).
A student solves \(x^2 - 5x = 0\) by dividing both sides by x and gets \(x = 5\). What did they miss?
Factorise instead of dividing by x.
Dividing by \(x\) can lose the solution \(x = 0\). Factorise: \(x(x - 5) = 0\), so \(x = 0\) or \(x = 5\).
Solve: \(x^2 = 3x + 10\). Give both answers separated by a comma.
First rearrange so one side equals 0.
Rearrange: \(x^2 - 3x - 10 = 0\). Factorise: \((x - 5)(x + 2) = 0\). So \(x = 5\) or \(x = -2\).
Which formula is used to solve \(ax^2 + bx + c = 0\)?
The formula starts with \(-b\).
The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
Solve: \(x^2 - 2x - 15 = 0\). Give both answers separated by a comma.
Find two numbers that multiply to -15 and add to -2.
\(x^2 - 2x - 15 = (x - 5)(x + 3)\). So \(x = 5\) or \(x = -3\).
Games
Practise this topic with interactive games.
Games coming soon.
Frequently Asked Questions
What is a quadratic equation?
An equation with a squared term, usually in the form ax² + bx + c = 0.
How can I solve quadratics?
By factorising, completing the square or using the quadratic formula.
How do I check solutions?
Substitute answers back into the original equation.