Solve quadratic equations step-by-step using five methods: quadratic formula, factoring, completing the square, inequalities, and full property analysis. Perfect for GCSE and A-Level maths revision.
Step-by-step learning with explanations
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Discriminant Rules
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A quadratic equation is any equation that can be rearranged into the standard form ax² + bx + c = 0, where a, b, and c are constants and a ≠ 0. The word "quadratic" comes from the Latin "quadratus" meaning square, because the variable is squared.
Quadratic equations are a cornerstone of GCSE and A-Level maths. They describe parabolas — the U-shaped curves you see in everything from planetary orbits to the path of a kicked football.
There are several ways to find the roots (solutions) of a quadratic equation. Each has its own advantages:
The "old reliable" method. It works for every quadratic, even those with complex roots or messy decimals. It relies on identifying the coefficients a, b, and c.
b² - 4ac
Equations that won't factor easily
The fastest method if the roots are integers. You find two numbers that multiply to c and add to b (when a = 1).
AC method, Difference of Two Squares
Simple whole-number solutions
Transforms the equation into vertex form. This is the best method for finding the turning point (maximum or minimum) of the parabola.
(h, k)
Finding vertex and graph transformations
The discriminant, Δ = b² - 4ac, tells you how many solutions exist before you even solve the equation:
The parabola crosses the x-axis twice.
The parabola touches the x-axis once (the vertex).
The parabola never touches the x-axis.
Using the vertex form y = a(x - h)² + k, you can easily see how the graph moves:
| Change | Effect on Parabola | Example |
|---|---|---|
| + k | Moves the graph UP | y = x² + 2 |
| - k | Moves the graph DOWN | y = x² - 3 |
| (x - h) | Moves the graph RIGHT | y = (x - 2)² |
| (x + h) | Moves the graph LEFT | y = (x + 3)² |
| a > 1 | Makes it narrower (stretch) | y = 3x² |
| 0 < a < 1 | Makes it wider (compression) | y = 0.5x² |
| Negative a | Flips it upside down (n-shape) | y = -x² |
Pro Tip: The vertex is always at (h, k). Notice that the sign of h inside the brackets is reversed — this is a classic exam trap!
Quadratic equations have several "traps" that catch out even the best students. Watch out for these:
The formula starts with -b. If b is already negative (e.g., b = -5), then -b becomes +5. Also, be careful with the -4ac part when c is negative.
Always use brackets when substituting negative numbers: -(-5) ± √((-5)² - 4(1)(-6)).
You cannot use the formula or factor correctly if the equation is not in the form ax² + bx + c = 0. For example, x² + 5x = 6 must be rewritten as x² + 5x - 6 = 0.
Rearrange everything to one side so the other side is 0 before you start identifying a, b, and c.
A common error is to write x = -b ± (√(b² - 4ac) / 2a). This is wrong — the ENTIRE numerator (-b ± √Δ) must be divided by 2a.
Draw a long fraction bar that goes under both the -b and the square root.
Students often think Δ < 0 means "no solution." In GCSE, this is mostly true (no real solutions), but in A-Level, it means there are two complex solutions.
Check the requirements of your specific exam board. Usually, you should state "no real roots" or "two complex roots."
In y = a(x - h)² + k, the vertex is (h, k). Students often get the sign of h wrong because of the minus sign in the formula.
Remember: "Inside is opposite." y = (x - 3)² has a vertex at +3. y = (x + 2)² has a vertex at -2.
Master the different solving techniques with these step-by-step examples:
Step 1: Identify b = -5 and c = 6.
Step 2: Find two numbers that multiply to 6 and add to -5.
The numbers are -2 and -3.
Step 3: Write in factored form.
(x - 2)(x - 3) = 0
Step 4: Solve each bracket.
x - 2 = 0 → x = 2
x - 3 = 0 → x = 3
Step 1: Identify coefficients.
a = 1, b = 2, c = -8
Step 2: Calculate the discriminant Δ = b² - 4ac.
Δ = (2)² - 4(1)(-8) = 4 + 32 = 36
Step 3: Apply the formula x = (-b ± √Δ) / 2a.
x = (-2 ± √36) / 2(1) = (-2 ± 6) / 2
Step 4: Solve for both cases.
Case 1: (-2 + 6) / 2 = 4 / 2 = 2
Case 2: (-2 - 6) / 2 = -8 / 2 = -4
Step 1: Take the coefficient of x, which is -6.
Step 2: Halve it (-3) and square it (9).
Step 3: Add and subtract this number in the equation.
y = (x² - 6x + 9) - 9 + 11
Step 4: Factor the perfect square trinomial.
y = (x - 3)² + 2
Step 5: Identify the vertex (h, k).
The vertex is (3, 2).
Even if you make a calculation error, writing down "x = (-b ± √(b² - 4ac)) / 2a" can earn you method marks.
Don't jump straight to the formula. If a = 1, spend 10 seconds checking if there are factors of c that add to b. It saves time!
Calculate b² - 4ac separately. Examiners often give a specific mark just for finding the discriminant correctly.
When substituting negative numbers into b² or -4ac, always use brackets to avoid sign errors on your calculator.
Watch if the question asks for "exact form" (surds), "2 decimal places," or "significant figures."
If you have time, substitute your x values back into the original equation. If they don't equal zero, something went wrong.
To solve ax² + bx + c = 0, you can use the Quadratic Formula: x = (-b ± √(b² - 4ac)) / 2a. Alternatively, you can solve by factoring (finding two numbers that multiply to c and add to b) or by completing the square.
The discriminant (Δ) is the part of the quadratic formula under the square root: b² - 4ac. It tells you the nature of the roots: Δ > 0 means two real roots, Δ = 0 means one repeated real root, and Δ < 0 means two complex (imaginary) roots.
Use factoring if the roots are simple integers (usually when a=1 and you can easily find factors). Use the Quadratic Formula for everything else, especially if the equation has decimals, large numbers, or if factoring is taking too long.
1. Ensure a = 1. 2. Move the constant c to the other side. 3. Take half of b, square it, and add it to both sides. 4. Factor the left side as a perfect square (x + b/2)². 5. Take the square root of both sides and solve for x.
Complex roots occur when the discriminant is negative (Δ < 0). Since you cannot take the square root of a negative number in the real number system, the solutions involve "i" (the imaginary unit, where i = √-1).
The x-coordinate of the vertex is x = -b / 2a. Substitute this x-value back into the original equation to find the y-coordinate. If the equation is in vertex form y = a(x - h)² + k, the vertex is simply (h, k).
When a ≠ 1, multiply a by c. Find two numbers that multiply to (a×c) and add to b. Split the middle term (bx) into these two numbers, then factor by grouping the first two terms and the last two terms.
First, solve the equation ax² + bx + c = 0 to find the critical values (roots). Then, sketch the parabola. For > 0, find where the graph is above the x-axis. For < 0, find where it is below. Express the result in interval notation.
Every quadratic equation has solutions, but they might not be "real" solutions. If Δ < 0, there are no real solutions, only complex ones. A graph of such an equation never crosses the x-axis.
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