Quadratic Equations
A quadratic equation contains a squared term such as x2. In this lesson you will turn quadratics into a form you can solve, then choose an efficient method and check your solutions.
Foundation & Higher: begin by solving suitable quadratics by factorising. Higher: extend to quadratics with a leading coefficient, the quadratic formula, completing the square and the number of real roots.
You should already be able to expand and factorise quadratic expressions. Revisit Factorising if you need a reminder.
Recognising a quadratic equation
A quadratic equation can be written in the form ax2 + bx + c = 0, where a is not zero. The powers of x are whole numbers and the highest power is 2.
For example, x2 + 5x + 6 = 0 is quadratic. The equation 3x + 7 = 0 is linear, while x3 − 2 = 0 is cubic.
Before solving, it is often useful to move every term to one side so the other side is zero. This makes factorising and the quadratic formula easier to use.
Solve by factorising
If a quadratic factorises, the zero-product rule turns one equation into two simple equations. If AB = 0, then A = 0 or B = 0.
Worked example — x² + 5x + 6 = 0
Factorise: x2 + 5x + 6 = (x + 2)(x + 3).
So (x + 2)(x + 3) = 0. Therefore x + 2 = 0 or x + 3 = 0.
x = −2 or x = −3.
Higher route — move terms first
Solve x2 + 7x = 18.
Move 18 to the left: x2 + 7x − 18 = 0.
Factorise: (x + 9)(x − 2) = 0, so x = −9 or x = 2.
Worked example — Interpret a contextual solution
A rectangle has sides x cm and (x + 2) cm and area 15 cm². The model is x(x + 2) = 15, so x² + 2x − 15 = 0 = (x + 5)(x − 3). The algebra gives x = −5 or 3. Reject −5 because a length cannot be negative: the sides are 3 cm and 5 cm, giving area 15 cm².
This contextual monic work is accessible on OCR Foundation. AQA places quadratic rearrangement in Higher; Pearson’s shared factorising target is x² + bx + c = 0. Use the corresponding route for your board.
Repeated roots and checking
Sometimes both factors give the same solution. For x2 − 6x + 9 = 0, factorising gives (x − 3)2 = 0, so x = 3. This is called a repeated root.
Check a proposed root by substituting it into the original equation. For x = 3: 32 − 6(3) + 9 = 9 − 18 + 9 = 0, so the solution checks.
Higher — factorising when a is not 1
Some Higher quadratics begin with a multiple of x2. The factorising method still works, but the first terms of the brackets must multiply to the leading term.
Worked example — 2x² + 7x + 3 = 0
Factorise: 2x2 + 7x + 3 = (2x + 1)(x + 3).
Set each factor to zero: 2x + 1 = 0 or x + 3 = 0.
x = −1/2 or x = −3.
Higher — the quadratic formula
The quadratic formula solves any equation written as ax2 + bx + c = 0:
Read a, b and c directly from the equation, including their signs.
Worked example — x² + 4x − 1 = 0
Here a = 1, b = 4 and c = −1.
Substitute into the formula: x = (−4 ± √(4² − 4 × 1 × −1)) ÷ 2 = (−4 ± √20) ÷ 2.
Simplify √20 = 2√5, giving x = −2 ± √5. To three decimal places, x ≈ 0.236 or x ≈ −4.236.
Worked example — keep the sign of b
For 3x2 − 5x − 2 = 0, a = 3, b = −5 and c = −2. Substitution gives x = (5 ± 7) ÷ 6, so x = 2 or x = −1/3.
Higher — completing the square
Completing the square rewrites a quadratic so the squared part is visible. For x² + bx, halve b to choose the number inside the square, because (x + p)² = x² + 2px + p². Subtract the extra p² to preserve the original expression. This can solve an equation and also connects naturally to its graph.
Worked example — x² + 6x + 1 = 0
Rewrite x2 + 6x as (x + 3)2 − 9.
So x2 + 6x + 1 = (x + 3)2 − 8.
Set equal to zero: (x + 3)2 = 8.
Take both square roots: x + 3 = ±√8 = ±2√2.
x = −3 ± 2√2.
Higher — how many real roots?
In the quadratic formula, the expression b2 − 4ac is called the discriminant. It tells you how many real roots there are.
b² − 4ac > 0: two distinct real roots. b² − 4ac = 0: one repeated real root. b² − 4ac < 0: no real roots.
For x2 + 2x + 5 = 0, the discriminant is 22 − 4(1)(5) = −16, so there are no real solutions.
Higher — completing the square when the coefficient is not 1
For ax² + bx + c with a ≠ 0, first take a outside the x² and x terms. Complete the square inside those brackets, then multiply the correction by a. The constant c can stay outside. This works when a is positive or negative.
Worked example — Complete the square in 2x² + 6x + 1
Take out 2 from the first two terms: 2(x² + 3x) + 1.
Half of 3 is 3/2. Since (x + 3/2)² = x² + 3x + 9/4, subtract 9/4 inside the brackets:
2[(x + 3/2)² − 9/4] + 1.
Multiply both parts by 2: 2(x + 3/2)² − 9/2 + 1.
2x² + 6x + 1 = 2(x + 3/2)² − 7/2. The subtraction is 9/2, not 9/4, because the outer 2 also multiplies it.
Use the form to solve 2x² + 6x + 1 = 0
2(x + 3/2)² = 7/2. Divide by 2: (x + 3/2)² = 7/4.
Take both square roots: x + 3/2 = ±√7/2.
x = (−3 ± √7)/2. Expanding the completed form gives 2x² + 6x + 9/2 − 7/2 = 2x² + 6x + 1. For either root, the squared bracket is 7/4, so the original expression is 2 × 7/4 − 7/2 = 0.
A negative leading coefficient
−2x² + 8x − 3 = −2(x² − 4x) − 3 = −2[(x − 2)² − 4] − 3 = −2(x − 2)² + 5. The two negative signs give +8 before combining constants.
Higher practice — Complete the square in 2x² − 6x + 1, then solve 2x² − 6x + 1 = 0.
2(x² − 3x) + 1 = 2[(x − 3/2)² − 9/4] + 1.
Completed form: 2(x − 3/2)² − 7/2.
Set it equal to zero: (x − 3/2)² = 7/4, so x = (3 ± √7)/2. Substitution into the completed form gives 2 × 7/4 − 7/2 = 0 for both roots.
Higher practice — Complete the square in −3x² + 6x + 2.
−3(x² − 2x) + 2 = −3[(x − 1)² − 1] + 2.
−3(x − 1)² + 5. Expanding gives −3x² + 6x − 3 + 5 = −3x² + 6x + 2.
Think Like a Mathematician
Why can a quadratic have two answers when a linear equation usually has one?
A quadratic can cross the horizontal axis at two different x-values. Algebraically, factorising can produce two factors, and either factor can be zero.
The method you choose does not change the roots. It changes only the route you take to find them.
Common Mistakes
Forgetting to make one side zero: x² + 7x = 18 is not ready to factorise as an equation until it becomes x² + 7x − 18 = 0.
Stopping after factorising: (x + 2)(x + 3) = 0 is not the final answer. Solve each factor.
Losing the ± sign: when taking a square root during completing the square, include both positive and negative roots.
Using the wrong b: in 3x² − 5x − 2 = 0, b is −5, not 5.
Rounding too early: keep exact roots such as −2 ± √5 until a decimal accuracy is requested.
Practice — solve the quadratics
Questions 1, 2, 4 and 5 use shared monic factorising in zero form. Question 3 requires rearrangement and follows the Higher route for AQA/Pearson. Questions 6–10 and 12 use Higher methods. Question 11 is a contextual monic case accessible on OCR Foundation; AQA places quadratics requiring rearrangement in Higher, and Pearson’s shared target is the standard zero form.
1. Solve x2 + 7x + 12 = 0.
Factorise: (x + 3)(x + 4) = 0.
x = −3 or x = −4.
2. Solve x2 − 5x + 6 = 0.
Factorise: (x − 2)(x − 3) = 0.
x = 2 or x = 3.
3. Solve x2 = 9x − 20.
Move everything left: x² − 9x + 20 = 0.
Factorise: (x − 4)(x − 5) = 0, so x = 4 or x = 5.
4. Solve x2 + 2x − 15 = 0.
Factorise: (x + 5)(x − 3) = 0.
x = −5 or x = 3.
5. Solve x2 − 10x + 25 = 0.
(x − 5)² = 0, so there is one repeated root: x = 5.
6. Solve 2x2 + 5x + 2 = 0.
Factorise: (2x + 1)(x + 2) = 0.
x = −1/2 or x = −2.
7. Use the quadratic formula to solve x2 + 2x − 7 = 0. Give exact answers.
a = 1, b = 2, c = −7.
x = (−2 ± √(4 + 28)) ÷ 2 = (−2 ± √32) ÷ 2.
√32 = 4√2, so x = −1 ± 2√2.
8. Use the quadratic formula to solve 3x2 − 4x − 2 = 0. Give exact answers.
a = 3, b = −4, c = −2.
x = (4 ± √(16 + 24)) ÷ 6 = (4 ± √40) ÷ 6.
√40 = 2√10, so x = (2 ± √10) / 3.
9. Solve x2 + 8x + 7 = 0 by completing the square.
x² + 8x + 7 = (x + 4)² − 9.
(x + 4)² = 9, so x + 4 = ±3.
x = −1 or x = −7.
10. Show that x2 + 2x + 5 = 0 has no real roots.
The discriminant is b² − 4ac = 2² − 4(1)(5) = 4 − 20 = −16.
It is negative, so there are no real roots.
11. A rectangle has sides x cm and (x + 3) cm. Its area is 40 cm². Find the positive value of x.
x(x + 3) = 40, so x² + 3x − 40 = 0.
Factorise: (x + 8)(x − 5) = 0, giving x = −8 or x = 5.
A length must be positive, so x = 5 cm.
12. Solve 4x2 − 12x + 9 = 0 and state what is special about the root.
4x² − 12x + 9 = (2x − 3)².
(2x − 3)² = 0, so x = 3/2. It is a repeated root.
Frequently Asked Questions
Should I always use the quadratic formula?
No. If a quadratic factorises neatly, factorising is usually quicker and clearer. The formula is especially useful when the factors are not obvious.
Why are there sometimes two answers?
A quadratic can be zero at two different x-values. The two roots are the x-values where its graph meets the x-axis.
What does a repeated root mean?
Both solutions have the same value. Algebraically the quadratic contains the same factor twice, such as (x − 3)².
When should I give decimals?
Keep exact answers involving fractions or surds unless the question asks for a stated decimal accuracy. If you do round, round only at the end.