A quadratic equation is an equation in which the highest power of the unknown x is 2; tidied up, it takes the form ax² + bx + c = 0 (a ≠ 0).
Just as finding when a ball thrown upwards is at height 0 gives both the moment it is thrown and the moment it lands, a quadratic equation usually has two solutions.
If the left-hand side can be factorised, it is solved from the fact that a product of two linear factors is 0; otherwise completing the square or the quadratic formula always works.
On a graph, the solutions are the x-coordinates where the parabola y = x² + bx + c meets the x-axis, and the number of meeting points is decided by the sign of the discriminant D = b² − 4ac.
Roots are where the parabola meets the x-axis
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Where the parabola y = x² + bx + c meets the x-axis, and the discriminant, as b and c change
-2
-3
👀 Roots on the graph
①If the parabola meets the x-axis at two points, there are two different roots (D > 0).
②If it touches the x-axis at one point, the two roots are equal: a repeated root (D = 0).
③If it does not meet the x-axis, there are no real roots (D < 0).
④At the starting values b = −2, c = −3, D = 16 and the roots are 3 and −1. When D is not a perfect square the roots are irrational, so their exact form is also written at the top of the picture.
Solving by factorising
Solving by factorising
ax² + bx + c = 0 → a(x − p)(x − q) = 0 → x = p or x = q
If a product of two numbers is 0, at least one of them is 0.
🧩 Steps for solving by factorising
①Move every term to the left so that the right-hand side is 0.
②Factorise the left-hand side.
③Set each factor equal to 0 and find x.
④For example, x² − 5x + 6 = 0 gives (x − 2)(x − 3) = 0, so x = 2 or x = 3.
Completing the square and square roots
Solving with square roots
(x + a)² = k (k ≥ 0) → x + a = ±√k → x = −a ± √k
Turn the left-hand side into a perfect square, then take square roots of both sides.
💡 Completing the square
①In x² + 6x = 7, add 9, the square of half the coefficient 6, to both sides: x² + 6x + 9 = 16.
②The left-hand side becomes (x + 3)², so (x + 3)² = 16.
③x + 3 = ±4.
④So x = 1 or x = −7.
The quadratic formula
Quadratic formula
x = −b ± √(b²−4ac)2a
It is the result of completing the square with general a, b and c, so it works for every quadratic equation.
Discriminant
D = b² − 4ac
D > 0: two different roots; D = 0: a repeated root; D < 0: no real roots.
⚠️ Care points with the formula
①Always rearrange to ax² + bx + c = 0 before reading off a, b and c.
②In −b at the front of the formula, the sign flips if b is negative.
③Divide the whole numerator by 2a; dividing only part of it is wrong.
④If D < 0, answer that there are no real roots.
⑤When a = 1 and b, c are integers, a perfect-square D means both roots are integers.
Worked examples
Example 1
Solve x²−5x+6=0.
1
Factorise the left-hand side: the two numbers with product 6 and sum −5 are −2 and −3.
x²−5x+6 = (x−2)(x−3)
2
Set each factor equal to 0.
(x−2)(x−3)=0 ⇒ x=2 or x=3
▸ x = 2 or x = 3
When factorising works, it is the fastest method.
Example 2
Solve x²−4x+1=0 with the quadratic formula.
1
Put a = 1, b = −4, c = 1 into the quadratic formula.
x = (4±√(16−4))/2
2
√12 = 2√3, so cancel 2 from the numerator and denominator.
x = (4±2√3)/2 = 2±√3
▸ x = 2 ± √3
If no factorisation is visible, use the formula and simplify the square root first.
Exam checklist
Three ways to solve a quadratic equation
factorising → completing the square → quadratic formula
If a factorisation is visible it is the fastest route; if not, use the formula.
G9 school-exam type
If x²−6x+k=0 has a double root, what is the constant k?
①3
②6
③9
④12
⑤36
▸ ③ 9
1
A double root means the discriminant D=0.
D = (−6)²−4·1·k = 36−4k
2
Solve 36−4k=0.
36−4k=0 ⇒ k=9
🎯 Exam points
①Try factorising first; if that fails, use the quadratic formula.
②The condition for a repeated root is D = b² − 4ac = 0.
③The sum of the roots is −b/a and their product is c/a.
④When asked for integer roots, check whether D is a perfect square.
⑤Always make the right-hand side 0 before starting.