01 · Keep every root
3x² = 12x
Hint
Move everything to one side and factor out 3x.
Worked solution
3x(x − 4) = 0
x = 0 or x = 4
Dividing by x would lose zero.
Understand · explore · practise
Solving quadratic equations
Solve quadratics by factorising, taking square roots and using the formula. Recognise hidden quadratics and check for lost or extra solutions.
Before you startExpanding, factorising and exact square roots
01 / Factorise
Make one side zero first.
A quadratic equation can be written as ax² + bx + c = 0, where a ≠ 0. Its highest power of x is 2. A solution makes the two sides equal.
AB = 0 ⇒ A = 0 or B = 0
A product is zero exactly when at least one factor is zero.
x² − x − 6 = 0
(x + 2)(x − 3) = 0
x = −2 or x = 3
For a non-unit leading coefficient, solve each linear factor:
2x² + 5x − 3 = 0
(2x − 1)(x + 3) = 0
x = 1/2 or x = −3
This rule depends on the product being zero. From (x + 2)(x − 3) = 10 you cannot set each factor equal to 10.
When does the product vanish?Try a value
(x + 2)(x − 3) = 0
Either x + 2 = 0 or x − 3 = 0.
At x = −2: 0 × (−5) = 0.
One zero factor is enough. The other factor need not be zero.
Watch a zero factor make the product zero
Pause, replay or seek freely. The notes explain the same idea and stay in view.
02 / Square roots
A squared bracket can be a shortcut.
If the equation already contains one squared bracket, isolate it and take both square roots. Keep irrational answers exact unless a decimal is requested.
(x − h)² = k
x = h ± √k when k > 0
When k = 0 there is one repeated root, x = h. When k < 0 there are no real solutions: a real square cannot be negative.
The symbol √11 names one non-negative number. The ± appears because you are solving an equation whose square could come from either sign.
Do not divide away a solution
x² = 7x
x(x − 7) = 0
x = 0 or x = 7
Dividing by x would assume x ≠ 0 and lose the solution x = 0. Factor out x instead.
Both signs surviveWorked example
(3x + 1)² = 11
The square is already isolated.
3x + 1 = ±√11
Use the positive and negative square roots.
x = (−1 ± √11)/3
Subtract 1, then divide the whole numerator by 3.
(3x + 1)² = (±√11)² = 11
Both answers pass the check.
03 / The formula
One method for every quadratic.
First collect all terms on one side. Read a, b and c with their signs from ax² + bx + c = 0.
x = [−b ± √(b² − 4ac)] / (2a)
a ≠ 0. Real solutions require b² − 4ac ≥ 0.
Put negative coefficients in brackets when squaring or multiplying. The denominator 2a divides both terms in the numerator.
Use the exact square-root expression until the final line. For 3x² + 2x − 7 = 0, the roots to 3 significant figures are 1.23 and −1.90.
A negative number under the root is useful information, not a calculator fault. For 2x² + 4x + 5 = 0, b² − 4ac = −24, so there are no real roots.
A quadratic that does not factorise neatlyWorked example
3x² + 2x − 7 = 0
a = 3, b = 2, c = −7.
b² − 4ac = 4 + 84 = 88
Keep the sign of c: subtracting 4 × 3 × (−7) adds 84.
x = (−2 ± √88)/6
Substitute into the formula.
x = (−1 ± √22)/3
√88 = 2√22; cancel a factor of 2 from the entire numerator and denominator.
04 / Choose a method
Let the form do some of the work.
- Visible factors and zero on the other side: use the zero-product rule.
- A squared bracket: isolate the square and take both roots.
- Simple factors available: factorise, then solve.
- Awkward coefficients: the formula is reliable.
- A turning point or maximum is also needed: completing the square gives that information as well as the roots.
Different methods must give the same solution set. Check an answer by substituting into the original equation.
Where does the formula come from?
For a ≠ 0, divide by a, move the constant and complete the square:
x² + (b/a)x = −c/a
(x + b/(2a))² = (b² − 4ac)/(4a²)
Multiplying by 4a² avoids any ambiguity about the sign of a when taking a square root.
(2ax + b)² = b² − 4ac
2ax + b = ±√(b² − 4ac)
x = [−b ± √(b² − 4ac)]/(2a)
This derivation assumes a real square root exists; otherwise there are no real roots.
Complete the square to solveWorked example
2x² + 4x − 5 = 0
Divide the whole equation by 2.
x² + 2x = 5/2
Move the constant.
(x + 1)² = 7/2
Add 1 to both sides.
x = −1 ± √14/2
Take both roots, then subtract 1.
05 / Hidden quadratics
Solve for the repeated expression first.
Look for a quantity and its square. Rename the quantity u, solve the quadratic in u, then return to x. The possible values of u matter.
Even powers: there can be four real roots
x⁴ − 13x² + 36 = 0
u = x² ≥ 0
(u − 4)(u − 9) = 0
x² = 4 or x² = 9
x = −3, −2, 2, 3
Odd powers: keep negative solutions
x⁶ − 7x³ − 8 = 0
u = x³
(u − 8)(u + 1) = 0
x = 2 or x = −1
An odd power takes both positive and negative real values. Each real cube has one real cube root.
Square roots: reject a negative u
x − 2√x − 8 = 0
u = √x ≥ 0
(u − 4)(u + 2) = 0
u = 4, so x = 16
u = −2 is impossible because √x is non-negative. Squaring −2 and accepting x = 4 would introduce a false solution.
Fractional powers with a cube root
x2/3 − x1/3 − 12 = 0
u = ∛x, so (u − 4)(u + 3) = 0
x = 4³ or x = (−3)³
x = 64 or x = −27
Here x2/3 means (∛x)², so negative real x is allowed.
A quadratic in an exponentialWorked example
42x − 17·4x + 16 = 0
Since 4²ˣ = (4ˣ)², let u = 4ˣ. Notice u > 0.
u² − 17u + 16 = 0
Factor the quadratic in u.
(u − 1)(u − 16) = 0
Both u = 1 and u = 16 are allowed.
4x = 1 or 4x = 16
x = 0 or x = 2
Recognise powers of 4. The answers here need no logarithms.
06 / Check restrictions
Rearranging can change the candidate list.
Record excluded values before clearing denominators. After squaring an equation, test every candidate in the original.
A rational equation
1/(x − 1) + 1/(x + 1) = 3/4
x ≠ 1 and x ≠ −1. Multiply by 4(x − 1)(x + 1):
4(x + 1) + 4(x − 1) = 3(x² − 1)
3x² − 8x − 3 = 0
(3x + 1)(x − 3) = 0
x = −1/3 or x = 3
Neither candidate is excluded; each makes the original left-hand side 3/4.
The restriction belongs to the original problem. It does not disappear when you cancel a factor or multiply out a denominator.
Squaring needs a final checkWorked example
√(2x + 3) = x
The left side is non-negative, so x must be non-negative too.
2x + 3 = x²
Squaring gives a necessary condition.
(x − 3)(x + 1) = 0
Candidates are 3 and −1.
x = 3: √9 = 3 ✓
x = −1: √1 ≠ −1
Only x = 3 solves the original equation.
07 / Your turn
Choose, solve, then check.
Use the simplest method you can justify. All roots requested here are real. Keep answers exact unless told otherwise.
02 · Use the formula
4x² − 4x − 5 = 0
Hint
a = 4, b = −4, c = −5. Put b in brackets when squaring.
Worked solution
D = (−4)² − 4(4)(−5) = 96
x = (4 ± √96)/8
x = (1 ± √6)/2
03 · Take square roots
(2x − 5)² = 13
Hint
Take both roots before isolating x.
Worked solution
2x − 5 = ±√13
x = (5 ± √13)/2
04 · An odd-power substitution
x⁶ − 26x³ − 27 = 0
Hint
Set u = x³ and factor the quadratic in u.
Worked solution
(u − 27)(u + 1) = 0
x³ = 27 or x³ = −1
x = 3 or x = −1
There are two real roots here; taking a cube root does not introduce a ±.
05 · A repeated expression
x − 5√x + 4 = 0
Hint
Set u = √x, with u ≥ 0.
Worked solution
u² − 5u + 4 = 0
(u − 1)(u − 4) = 0
x = 1² or x = 4²
x = 1 or x = 16
Substitution gives 1 − 5 + 4 = 0 and 16 − 20 + 4 = 0.
06 · Spot the extra solution
√(x + 6) = x
Hint
x must be non-negative. Squaring can introduce a negative candidate.
Worked solution
x + 6 = x²
(x − 3)(x + 2) = 0
x = 3 passes: √9 = 3. x = −2 fails: √4 = 2, not −2. The only solution is x = 3.
07 · A denominator to remember
1/x + 1/(x + 3) = 2/3
Hint
Exclude x = 0 and x = −3 before multiplying by 3x(x + 3).
Worked solution
3(x + 3) + 3x = 2x(x + 3)
2x² = 9
x = ±3√2/2
Both roots are allowed. The common-denominator numerator is 2x + 3; substituting either root gives 2/3.
08 / Recap
Keep the original equation in view.
- Factorise only after making one side zero.
- A squared bracket needs both square-root signs, unless its value is zero.
- In the formula, keep coefficient signs and divide the whole numerator by 2a.
- For a hidden quadratic, solve in u and then finish solving in x.
- Keep denominator and square-root restrictions. Check candidates after squaring.
Section 1 of 8 · Factorise