A quadratic equation has the form ax² + bx + c = 0, where a is not 0, so the highest power of the variable is 2. Free notes and 12 questions answered.
Updated 25 Mar 2026 · Revise Right Team
Solving Quadratics, Simultaneous Equations and Rearranging Formulae is part of Solving equations and inequalities in GCSE AQA Higher Maths.
4 min
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GCSE AQA Higher
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What this lesson covers
A quadratic equation has the form ax² + bx + c = 0, where a is not 0, so the highest power of the variable is 2. Free notes and 12 questions answered.
Solve quadratic equations using multiple methods including factorisation, completing the square, and the quadratic formula
Apply the discriminant to determine the nature and number of roots of quadratic equations
Solve simultaneous equations involving one linear and one quadratic equation
Rearrange complex formulae to make any variable the subject
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Use this lesson to stay inside Solving equations and inequalities while connecting the detail back to the wider Algebra topic.
Discriminant: The expression b² - 4ac from the quadratic formula that determines the nature of roots. When positive, there are two real roots; when zero, one repeated root; when negative, no real roots.
Completing the square: A method for solving quadratic equations by rewriting ax² + bx + c = 0 in the form a(x + p)² + q = 0, making the equation easier to solve.
Simultaneous equations: A system of equations that must be satisfied at the same time. Solutions are the values that make all equations true simultaneously.
Subject of formula: The variable that appears alone on one side of an equation, with all other variables and constants on the other side.
Revision notes
Read the core explanations from Solving Quadratics, Simultaneous Equations and Rearranging Formulae before testing yourself from memory.
1. Beyond Basic Algebra
You've mastered linear equations like $2x + 3 = 11$. But GCSE throws curveballs: equations with $x^2$ terms, fractions that make your head spin, and formulae where you need to rearrange for different variables.
These advanced methods aren't just exam tricks. Engineers use quadratic equations to design bridges. Scientists rearrange formulae to isolate unknown quantities. The discriminant tells us whether a projectile will hit its target.
Here's the thing: each method has its sweet spot. Factorisation works beautifully when numbers are friendly. The quadratic formula handles anything, but it's overkill for simple cases. Completing the square reveals the vertex of parabolas.
We'll build your toolkit systematically. Start with the patterns you can spot by eye, then move to the heavy artillery when things get messy. By the end, you'll know which weapon to draw for any equation.
2. The Quadratic Equation Arsenal
Quadratic equations come in the standard form $ax^2 + bx + c = 0$. The key insight: there are usually two solutions because a parabola typically crosses the x-axis at two points.
Your first weapon is factorisation. Look for two numbers that multiply to give $ac$ and add to give $b$. This works when the numbers are integers and relatively small.
Worked example: Solving by Factorisation
Solve $x^2 - 5x + 6 = 0$
Need two numbers that multiply to 6 and add to -5 — Look at the constant term and coefficient of x
$-2 \times -3 = 6$ and $-2 + (-3) = -5$ — Both conditions satisfied
$(x - 2)(x - 3) = 0$ — Each bracket contains one of our numbers
$x - 2 = 0 \Rightarrow x = 2$
$x - 3 = 0 \Rightarrow x = 3$ — If a product equals zero, at least one factor must be zero
Answer: $x = 2$ or $x = 3$
When factorisation fails, completing the square becomes your ally. This method transforms the equation into a perfect square, making it solvable by taking square roots.
Worked example: Completing the Square Method
Solve $x^2 + 6x + 5 = 0$ by completing the square
Half of 6 is 3 — This will be the number inside our bracket
$x = -3 + 2 = -1$ or $x = -3 - 2 = -5$ — Two solutions from the ± symbol
Answer: $x = -1$ or $x = -5$
The quadratic formula is your nuclear option. It works for any quadratic, even when the numbers are messy. The expression under the square root sign is called the discriminant.
3. Simultaneous Equations with Quadratics
When one equation is linear and the other quadratic, you're finding where a straight line intersects a curve. This may give zero, one, or two intersection points depending on the discriminant of the resulting quadratic.
The strategy is substitution. Use the linear equation to express one variable in terms of the other, then substitute into the quadratic equation. This eliminates one variable, leaving you with a quadratic to solve.
Worked example: Linear-Quadratic Simultaneous Equations
$5x^2 + 4x + 1 = 10$
$5x^2 + 4x - 9 = 0$ — Rearrange to standard form
$x = \frac{-4 \pm \sqrt{16 + 180}}{10} = \frac{-4 \pm \sqrt{196}}{10} = \frac{-4 \pm 14}{10}$ — Here a = 5, b = 4, c = -9
$x = \frac{-4 + 14}{10} = 1$ or $x = \frac{-4 - 14}{10} = -1.8$ — Two x-values from the ± in the formula
When $x = 1$: $y = 2(1) + 1 = 3$
When $x = -1.8$: $y = 2(-1.8) + 1 = -2.6$ — Substitute back into the linear equation
Answer: $(1, 3)$ and $(-1.8, -2.6)$
Notice how we got exactly two solutions. This makes geometric sense: a straight line typically intersects a circle at two points. If the discriminant had been zero, the line would be tangent to the circle (one intersection point).
This block has been removed. Please refer to other parts of the lesson for detailed examples.
4. Advanced Formula Rearrangement
Making a different variable the subject of a formula is like solving an equation, but with letters instead of numbers. The same rules apply: whatever you do to one side, do to the other.
The challenge comes when your target variable appears multiple times, or when it's trapped inside square roots, fractions, or powers. You need a systematic approach to isolate it.
Worked example: Rearranging Complex Formulae
Make $r$ the subject of $A = \pi r^2 + 2\pi rh$
$A = \pi r(r + 2h)$ — Both terms contain $\pi r$, so factor it out
$\frac{A}{\pi} = r(r + 2h)$ — Isolate the bracket containing r
$\frac{A}{\pi} = r^2 + 2hr$ — We now have a quadratic in r
$r^2 + 2hr - \frac{A}{\pi} = 0$ — Standard quadratic form: ar² + br + c = 0
$r = \frac{-2h \pm \sqrt{4h^2 + 4A/\pi}}{2}$ — Here a = 1, b = 2h, c = -A/π
$r = \frac{-2h \pm 2\sqrt{h^2 + A/\pi}}{2} = -h \pm \sqrt{h^2 + \frac{A}{\pi}}$ — Take positive root since r must be positive
Answer: $r = -h + \sqrt{h^2 + \frac{A}{\pi}}$
When the subject appears in a denominator, multiply through to clear fractions first. When it's under a square root, square both sides (but watch out for extraneous solutions).
Worked example: Subject Under a Square Root
Make $x$ the subject of $y = \sqrt{\frac{x + 3}{x - 1}}$
$y^2 = \frac{x + 3}{x - 1}$ — This eliminates the square root
$y^2(x - 1) = x + 3$ — Clear the fraction
$y^2x - y^2 = x + 3$ — Distribute the multiplication
$y^2x - x = y^2 + 3$
$x(y^2 - 1) = y^2 + 3$ — Factor out x from the left side
This block has been removed. Please refer to other parts of the lesson for fill-in-the-blank activities. As a replacement, consider additional derivation-related content.
5. Practice and Consolidation
Time to test your arsenal. These problems combine different techniques, so you'll need to choose the right method for each situation.
The key insight is recognising patterns. Perfect squares suggest completing the square or difference of squares. Messy coefficients point toward the quadratic formula. Multiple appearances of your target variable mean you might need to treat it as a quadratic.
6. Exam Strategy and Common Traps
Examiners love testing your method selection. They'll give you a quadratic that could be solved multiple ways, but one method is clearly easier. Spotting the efficient path saves precious exam time.
Watch for disguised quadratics. An equation like $\frac{1}{x} + \frac{1}{x+1} = \frac{1}{2}$ becomes quadratic when you clear the fractions.
For simultaneous equations, always check your solutions by substituting back into both original equations. A common error is solving correctly but making arithmetic mistakes in the back-substitution.
Remember that Higher tier questions often combine multiple techniques. You might need to rearrange a formula, then solve the resulting quadratic, then interpret the solutions in context. Practice identifying which method suits each equation type.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-03-25.
A quadratic equation is an equation of the form ax² + bx + c = 0, where a is not 0, so the highest power of the variable is 2. It can be solved by factorisation, completing the square or the quadratic formula.
How do you solve linear and quadratic simultaneous equations?
Substitute the linear equation into the quadratic one to eliminate a variable, then solve the resulting quadratic. For y = 2x + 1 and x² + y² = 10, x² + (2x + 1)² = 10 gives 5x² + 4x - 9 = 0, so x = 1 or x = -1.8, giving the solutions (1, 3) and (-1.8, -2.6).
Why do quadratic equations usually have two solutions?
Quadratic equations usually have two solutions because a parabola typically crosses the x-axis at two points. The discriminant b² - 4ac shows this: positive gives two real roots, zero gives one repeated root and negative gives no real roots.
How many times can a straight line meet a circle?
A straight line can meet a curve such as a circle at zero, one or two points, depending on the discriminant of the quadratic formed after substitution. If the discriminant is zero, the line is a tangent to the circle and there is exactly one intersection point.
How do you make x the subject when it is under a square root?
Square both sides, clear the fraction, collect the x terms, then factorise. For y = √((x + 3)/(x - 1)), squaring gives y²(x - 1) = x + 3, so y²x - x = y² + 3 and x = (y² + 3)/(y² - 1).
How do you rearrange a formula that becomes a quadratic?
If the subject appears more than once, including as a square, rearrange into a quadratic and use the quadratic formula. For A = πr² + 2πrh, r² + 2hr - A/π = 0, so r = -h + √(h² + A/π), taking the positive root since r must be positive.
What is a disguised quadratic?
A disguised quadratic is an equation that becomes quadratic once it is rearranged. For example, 1/x + 1/(x + 1) = 1/2 becomes a quadratic when the fractions are cleared, and it can then be solved by factorising or the quadratic formula.
How do you solve x² + 6x + 5 = 0 by completing the square?
x² + 6x + 5 = 0 is solved by completing the square: (x + 3)² - 9 + 5 = 0, so (x + 3)² = 4. Square-rooting gives x + 3 = ±2, so x = -1 or x = -5.
How do you solve y = 2x + 1 and x² + y² = 10?
y = 2x + 1 and x² + y² = 10 are solved by substituting: x² + (2x + 1)² = 10 gives 5x² + 4x - 9 = 0. The quadratic formula gives x = 1 or x = -1.8, so the solutions are (1, 3) and (-1.8, -2.6).
How do you make r the subject of A = πr² + 2πrh?
To make r the subject of A = πr² + 2πrh, divide by π and rearrange to the quadratic r² + 2hr - A/π = 0. The quadratic formula then gives r = -h + √(h² + A/π), taking the positive root since r must be positive.
How do you check linear and quadratic simultaneous solutions?
Solutions to linear and quadratic simultaneous equations are checked by substituting each pair back into both original equations. A common error is solving correctly but making arithmetic mistakes when substituting back to find y.
How do you deal with a square root when rearranging a formula?
A square root is removed when rearranging by squaring both sides, though extraneous solutions should be watched for. For y = √((x + 3)/(x - 1)), squaring gives y² = (x + 3)/(x - 1), and clearing the fraction leads to x = (y² + 3)/(y² - 1).