Algebra
Grade 5-7Quadratic Equations
Solving a quadratic equation means finding every value of x that makes it true, and for most GCSE quadratics this is done by factorising and setting each bracket equal to zero. This lesson covers solving equations of the form x² = a, solving equations with no constant term, solving by factorising when the coefficient of x² is 1, and solving when the coefficient of x² is greater than 1.
What you need to know
- Rearrange the equation into the form a x squared plus b x plus c equals 0 first.
- If it factorises, set each bracket equal to 0 to get the two solutions.
- A quadratic usually has two solutions, sometimes one repeated, sometimes none real.
- Never divide both sides by x; you would lose the solution x = 0.
Written by Asad, co-founder of Teachably
Video walkthrough coming soon
The written lesson below covers everything you need in the meantime.
Solving x² = a
Take the square root of both sides, remembering that there are two possible solutions, one positive and one negative.
To solve x² = 25, take the square root of both sides: x = ±√25, so x = 5 or x = −5.
Solving equations with no constant term
Factorise out the common factor of x, then set each bracket equal to zero separately, since a product is zero only when at least one of its factors is zero.
To solve x² + 5x = 0, factorise to get x(x + 5) = 0, so either x = 0 or x + 5 = 0, giving x = 0 or x = −5.
Solving by factorising when the coefficient of x² is 1
Factorise the quadratic into two brackets, then set each bracket equal to zero and solve each resulting equation.
To solve x² − 5x + 6 = 0, factorise to get (x − 2)(x − 3) = 0, so x − 2 = 0 or x − 3 = 0, giving x = 2 or x = 3.
Solving when the coefficient of x² is greater than 1
Factorise using the same method as for a quadratic with a larger coefficient, then set each bracket equal to zero and solve, taking care with any fractional solutions.
To solve 2x² + 7x + 3 = 0, factorise to get (2x + 1)(x + 3) = 0, so 2x + 1 = 0 or x + 3 = 0, giving x = −1/2 or x = −3.
Worked examples
Three exam-style questions, fully solved
Solve x² = 25.
Easy- 1.Take the square root of both sides: x = ±√25
Answer: x = 5 or x = −5
Solve x² − 5x + 6 = 0.
Medium- 1.Find two numbers that multiply to 6 and add to −5: −2 and −3
- 2.Factorise: (x − 2)(x − 3) = 0
- 3.Set each bracket equal to zero: x − 2 = 0 or x − 3 = 0
Answer: x = 2 or x = 3
Solve 2x² + 7x + 3 = 0.
Hard- 1.Multiply the coefficient of x² by the constant term: 2 × 3 = 6, then find two numbers that multiply to 6 and add to 7: 6 and 1
- 2.Split the middle term and factorise in pairs: (2x + 1)(x + 3) = 0
- 3.Set each bracket equal to zero: 2x + 1 = 0 or x + 3 = 0
Answer: x = −1/2 or x = −3
Practice
6 questions, marked instantly
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Practice
Now try these yourself.
Type your answer and check it. The worked solution appears once you have had a go.
Solve x² = 36.
Solve x² + 7x + 12 = 0.
Solve x² − 5x = 0.
Solve x² + 2x − 15 = 0.
Solve x² − 8x + 16 = 0.
Solve 2x² − 5x − 3 = 0.
Avoid these
The mistakes students make most often
When solving x² = a, only giving the positive square root and forgetting that the negative square root is also a solution.
Forgetting to rearrange the equation so it equals zero before factorising, especially when terms appear on both sides of the equals sign.
After factorising into two brackets, forgetting that each bracket must be set equal to zero separately, or making a sign error solving one of the resulting equations.
When the coefficient of x² is greater than 1, making an error dividing to find the solution, especially forgetting to include the coefficient of x in the denominator.
Not checking each solution by substituting it back into the original equation to confirm it works.
FAQ
Questions students and parents ask
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