Algebra · GCSE Maths
Iteration and trial and improvement
GCSE Maths iteration: trial and improvement to find roots of equations, iterative formulas xₙ₊₁ = f(xₙ), and when a solution converges or diverges.
Trial and improvement: pick a value, test too high or too low, narrow the interval. Iteration: substitute into the formula and repeat until the answer settles.
The important bits
What you need to know
- 1
Trial and improvement finds a solution to an equation such as x³ − 2x = 5 by testing values and narrowing to one decimal place.
- 2
Pick a starting value. If the result is too high, try a smaller x; if too low, try larger. Each trial must show the calculation, not just a guess.
- 3
To one decimal place, test until you have shown the answer is between two values that differ by 0.1, with the correct value on one side.
- 4
An iterative formula writes xₙ₊₁ = f(xₙ). Start with x₀, calculate x₁, then x₂, and continue until the value stops changing to the required accuracy.
- 5
A root of f(x) = 0 is where the graph crosses the x-axis. Trial and improvement and iteration both hunt that crossing without algebraically solving.
- 6
If an iteration converges, successive values get closer to the root. If it diverges, values shoot away — try a different starting value or formula.
- 7
Rearrange the equation into x = … form before iterating. x² − 3x + 1 = 0 might become x = (x² + 1)/3 depending on the formula given.
- 8
Always check your final answer in the original equation, not in the rearranged iterative form, which might have introduced extraneous solutions.
Quotations worth analysing
Short evidence. Real method.
“xₙ₊₁ = f(xₙ)”
Each new value comes from the previous one. Students who write x = f(x) once and stop have done one step, not an iteration.
“Too high, try lower; too low, try higher”
The direction of your next trial must follow the test. If x³ − 2x is greater than 5, your x was too large for this increasing region.
“Show the interval that traps the answer”
For x = 2.3 to 1 d.p., show f(2.25) and f(2.35) on opposite sides of zero, or equivalent bounds. A lone 2.3 with no interval loses the accuracy mark.
Go deeper
Trial and improvement step by step
Solve x³ − 2x = 5 to 1 decimal place. Try x = 2: 8 − 4 = 4, too low. Try x = 3: 27 − 6 = 21, too high. Try x = 2.5: 15.625 − 5 = 10.625, still high. Try 2.2: 10.648 − 4.4 = 6.248, high. Try 2.1: 9.261 − 4.2 = 5.061, slightly high. Try 2.05: 8.615 − 4.1 = 4.515, low. The root lies between 2.05 and 2.1. For 1 d.p., test 2.15: 9.94 − 4.3 = 5.64, high; 2.04: 8.49 − 4.08 = 4.41, low. So x = 2.1 to 1 d.p. because 2.1 gives 5.061 (close) and values below 2.05 are clearly too low. Write every trial as a line: x = …, calculation = …, comment too high/low. Examiners award method for the narrowing, not for psychic guessing.
Go deeper
When iteration converges
Given xₙ₊₁ = √(xₙ + 2) with x₀ = 1: x₁ = √3 ≈ 1.732, x₂ ≈ 1.932, x₃ ≈ 1.976, x₄ ≈ 1.994, x₅ ≈ 1.998. The values settle near 2. Check in x² − x − 2 = 0: 4 − 2 − 2 = 0. If the question asks for 2 decimal places, continue until two successive values agree. Divergence looks different: xₙ₊₁ = xₙ² − 2 with x₀ = 2 gives 2, 2, 2 if lucky, but x₀ = 3 gives 7, 47, 2207 — values explode. GCSE questions usually give a convergent formula. Draw a cobweb or staircase diagram if asked: the graph of y = x and y = f(x) crossing shows the fixed point. The iteration chases that intersection.
Go deeper
Rearranging before you iterate
The same equation can have several iterative forms. x² = x + 6 gives x = √(x + 6) or x = (x² − 6) depending on rearrangement — only some converge near a given root. x² − x − 6 = 0 has roots 3 and −2. √(x + 6) with x₀ = 2 converges to 3. The form x = (x² − 6)/x might not converge at all near 3. If a question gives the formula, use it. If you must choose, test two steps mentally: do values move toward a sensible root? Trial and improvement does not need rearrangement — it needs organised trials. Iteration needs the formula and a starting value. Do not mix the methods on the same line. Trial is bracket narrowing; iteration is repeated substitution.
See the idea in action
Show that x³ − x − 4 = 0 has a root between 1 and 2. Then find it to 1 decimal place. At x = 1: 1 − 1 − 4 = −4 (negative). At x = 2: 8 − 2 − 4 = 4 (positive). Sign change ⇒ root between 1 and 2. x = 1.5: 3.375 − 1.5 − 4 = −2.125 (negative). Root between 1.5 and 2. x = 1.7: 4.913 − 1.7 − 4 = −0.187 (negative). x = 1.8: 5.832 − 1.8 − 4 = 0.032 (positive). Root between 1.7 and 1.8. For 1 d.p.: x = 1.8 gives +0.032; x = 1.75 gives −0.081. So x = 1.8 to 1 d.p.
Exam technique
Turn knowledge into marks
Trial and improvement: every line needs x, the value of the expression, and too high/too low. Iteration: list x₀, x₁, x₂ until the change is smaller than the accuracy required.
Common mistakes
Do not give these marks away
- 01
Stopping after one iterative step when the question asks for convergence to 2 decimal places.
- 02
Giving an answer to 1 d.p. without showing the trapping interval (e.g. 2.15 too high, 2.05 too low).
- 03
Checking the answer in the rearranged formula instead of the original equation.
To solve an equation by trial and improvement to 1 decimal place, you must
AGuess once and box the answer
BShow trials that narrow the root into an interval of width 0.1
CUse the quadratic formula
DIterate at least ten times regardless of accuracy
Show the answer
Show trials that narrow the root into an interval of width 0.1. Trial and improvement is organised narrowing with shown calculations. One guess is not a method. Ten iterations without a trapping interval does not prove 1 d.p.
Quick questions
If this is the bit you searched
What is the difference between trial and improvement and iteration?
Trial and improvement tests values and narrows an interval. Iteration uses a formula xₙ₊₁ = f(xₙ) and repeats substitution until values settle.
How many iterations should I show?
Enough for the required accuracy: when successive values agree to the decimal places asked, or when the change is clearly negligible.
What if my iteration diverges?
Values grow without settling. Try a different starting value or a different rearrangement. GCSE papers usually give a convergent formula.
Why show a sign change first?
It proves a root exists in that interval (continuous function). It is often the first mark on “show there is a root between…” questions.