Solving an equation means finding the value(s) of the unknown that make the equation true. This is the culmination of all the algebra skills you’ve built — expressions, indices, expanding, factorising and rearranging all come together here.
Linear equationsQuadratic by factorisingQuadratic formulaThe discriminantSimultaneous equations
What does “solving” mean?
Finding the value(s) that make the equation true. In x² − 5x + 6 = 0, substituting x=2 gives 4 − 10 + 6 = 0 ✔ — that’s a solution!
Solution: a value that makes both sides equal
Root: same as solution (used for quadratics)
Check: always substitute back to verify!
💡 Always substitute your answer back into the original equation to check. This catches errors before they cost marks!
Which method to use?
Linear (no x²) — rearrange using inverse operations
Quadratic — can factorise — always try factorising first (faster)
Quadratic — won’t factorise — use the quadratic formula
Two equations, two unknowns — simultaneous equations
📌 In T Level exams, quadratic equations almost always factorise neatly. Try factorising first — if you can’t find integer pairs within 30 seconds, switch to the formula.
↔
Solving Linear Equations
A linear equation has no x² or higher powers — just x. The strategy is always the same: collect x terms on one side, numbers on the other, then divide.
Strategy: Expand brackets → collect x terms on one side → collect numbers on the other → divide to find x.
Equations with x on both sides
Solve: 5x − 3 = 2x + 9
Collect x: 5x − 2x = 9 + 3
Simplify: 3x = 12
Divide: x = 12/3
Answer:x = 4 Check: 5(4)−3=17, 2(4)+9=17 ✔
Equations with brackets
Solve: 3(2x − 1) = 2(x + 7)
Expand: 6x − 3 = 2x + 14
Collect x: 4x = 17
Answer:x = 17/4 = 4.25 ✅
⚙ Solve: 4(t − 2) = 3(t + 1) for time t
Expand: 4t − 8 = 3t + 3
Collect t: t = 11
Answer:t = 11 s ✅
⚠️ Most common error: Moving a term without changing its sign. 5x − 3 = 2x + 9 → 5x − 2x = 9 + 3 (the −3 becomes +3 when it crosses the = sign)
🧩
Solving Quadratics by Factorising
The fastest method when it works. Factorise the quadratic, then use the zero product rule: if two things multiply to give zero, at least one of them must be zero.
Zero product rule: If A × B = 0, then A = 0 OR B = 0 (or both). This is the key that unlocks quadratic equations!
Step-by-step method
Rearrange so everything is on one side: ax² + bx + c = 0
Factorise the left side completely
Set each bracket equal to zero: bracket = 0
Solve each simple equation to find x
Check both answers by substituting back in
✅ A quadratic can have 2, 1 or 0 real solutions. The discriminant (next tab) tells you which.
More worked examples
Solve: x² − 9x + 20 = 0
Factorise: need ×=20, +=−9 → −4 and −5
Brackets: (x − 4)(x − 5) = 0
Solve:x = 4 or x = 5 ✅
Solve: x² − 4x = 0 (no constant!)
Factorise: x(x − 4) = 0
Solve: x = 0 or x − 4 = 0
Answer:x = 0 or x = 4 ✅
🧮
The Quadratic Formula
When factorising is difficult or impossible, the quadratic formula always works. It gives the exact solutions to any quadratic equation — even ones with decimal or irrational roots.
How to use it
Write in the form ax² + bx + c = 0 and read off a, b, c
Calculate the discriminant: b² − 4ac
If discriminant < 0: no real solutions. If ≥ 0: continue
Substitute into formula — calculate + version and − version separately
Simplify each answer
⚠️ The ± sign gives TWO calculations. Do them separately! One uses +√ (larger root), the other uses −√ (smaller root).
Worked: 2x² + 5x − 3 = 0
Solve: 2x² + 5x − 3 = 0
a=2, b=5, c=−3
b²−4ac: 25 − 4(2)(−3) = 25 + 24 = 49
√49 = 7
x = (+): (−5 + 7) / 4 = 2/4 = 0.5
x = (−): (−5 − 7) / 4 = −12/4 = −3
Answer:x = 0.5 or x = −3 ✅
Solve: x² − 6x + 7 = 0 (won’t factorise neatly — use the formula)
a=1, b=−6, c=7
Discriminant: 36 − 28 = 8
√8 = 2√2 ≈ 2.828
x = (+): (6 + 2.828)/2 = 4.414
x = (−): (6 − 2.828)/2 = 1.586
Answer:x = 4.41 or x = 1.59 (3 s.f.) ✅
△
The Discriminant
The expression b² − 4ac inside the square root tells you how many real solutions a quadratic has — before you even solve it. It’s a powerful tool for checking and engineering design.
Using the discriminant
b²−4ac < 0 → 0 real roots (no solution)
b²−4ac = 0 → 1 repeated root
b²−4ac > 0 → 2 distinct real roots
📌 In engineering this tells you whether a system has a physically meaningful solution before you waste time solving it fully.
Worked Examples
How many solutions has: x² − 4x + 5 = 0?
a=1,b=−4,c=5
Discriminant: (−4)² − 4(1)(5) = 16 − 20 = −4
Answer:−4 < 0 → No real solutions ✅
Find k so that x² + kx + 9 = 0 has one solution
Condition: b² − 4ac = 0
Substitute: k² − 36 = 0
Solve:k = ±6 ✅
⚡
Complex Factorising Problems
Harder quadratics where a ≠ 1, expressions that need rearranging first, equations disguised as higher powers, and quadratics in disguise. These are the T Level challenge problems.
Type 1 — Rearrange firstAlways get to ax²+bx+c=0 before factorising
Type 4 — Quadratic in disguiseSubstitute u = x² to reveal hidden quadratic
Solve: x⁴ − 5x² + 4 = 0
Let u=x²: u² − 5u + 4 = 0
Factorise u: (u−1)(u−4) = 0
u=1 or u=4
x²=1: x = ±1
x²=4:x = ±2 All 4 roots: x=1,−1,2,−2 ✅
Type 5 — Harder ax²+bx+c using AC method6x² + 11x − 10 = 0
Full AC method: 6x² + 11x − 10 = 0
a=6,b=11,c=−10 ac = −60
Find p,q: p×q=−60, p+q=11 → 15 and −4
Split 11x: 6x² + 15x − 4x − 10 = 0
Group: 3x(2x+5) − 2(2x+5) = 0
Factor out: (2x+5)(3x−2) = 0
Answer:x = −2.5 or x = 2/3 ✅
⚙ Engineering context: 6x² + 11x − 10 = 0 where x = dimension in mm
Roots: x = −2.5 or x = 2/3
Physical: dimension can’t be negative
Discard: x = −2.5
Answer:x = 2/3 mm ≈ 0.667 mm ✅
🔀
Simultaneous Equations
Two equations, two unknowns. Solve both at the same time using elimination or substitution. Essential for circuit analysis, force balance and any problem with two related unknowns.
Method 1 — EliminationAdd or subtract equations to remove one variable
Match coefficients — multiply equations so one variable has equal coefficients
Add or subtract the equations to eliminate that variable
Solve the resulting single equation
Substitute back to find the second variable
Check in BOTH original equations
Solve: 3x + 2y = 16 and x + 2y = 8
Subtract: (3x+2y) − (x+2y) = 16−8
Simplify: 2x = 8 → x = 4
Substitute: 4 + 2y = 8 → 2y = 4 → y = 2
Answer:x = 4, y = 2 ✅
Method 2 — SubstitutionRearrange one equation and substitute into the other
Rearrange one equation to express one variable in terms of the other
Substitute this expression into the other equation
Solve for the remaining variable
Substitute back to find the first variable
Solve: y = 2x + 1 and 3x + y = 16
Substitute: 3x + (2x+1) = 16
Simplify: 5x + 1 = 16 → 5x = 15 → x = 3
Find y: y = 2(3)+1 = 7
Answer:x = 3, y = 7 ✅
⚙ Engineering: Two-loop circuit — Kirchhoff’s voltage law gives two equations