Trial and Improvement: KS3 Maths Step-by-Step

Trial and improvement is a systematic way of finding approximate solutions to equations that are too complex to solve algebraically. You substitute a guess, compare the result to the target, then adjust and repeat — each time narrowing the interval until you reach the required degree of accuracy.

When do you use trial and improvement?

Trial and improvement is useful whenever an equation cannot be solved by straightforward rearranging or factorising. At KS3, you will most commonly meet it when the equation contains x³ (or a combination such as x³ + x, or x³ − 2x) set equal to a target value. Typical examples include:

x³ = 50

x³ + x = 20

x³ − 2x = 10

These equations have solutions that are not whole numbers or simple fractions, so standard algebraic methods will not give you an exact answer. Instead, you close in on the solution by repeatedly substituting trial values, comparing each result to the target, and adjusting until you have narrowed things down sufficiently.

Equation type | Can you solve algebraically? | Approach
3x + 5 = 14 | Yes — rearrange | Rearrange: x = 3
x² − 5x + 6 = 0 | Yes — factorise | Factorise: (x − 2)(x − 3) = 0
x³ + x = 20 | No | Trial and improvement
x³ − 3x = 8 | No | Trial and improvement

Trial and improvement typically asks for the answer correct to 1 decimal place (1 d.p.), though some questions ask for 2 d.p. Your method must be clear, systematic, and fully recorded so the examiner can see every trial you make.

How does the trial and improvement method work?

The method follows a logical five-step sequence. Mastering this sequence means you will always know what to do next, regardless of the equation.

Find an initial bracket. Substitute integer values into the equation until you find one integer whose result is too small and one whose result is too big. The solution lies between those two integers.

Test the midpoint of the bracket. If your bracket is (2, 3), try x = 2.5. This tells you which half of the bracket holds the solution.

Narrow the bracket. Test further values between the two closest bounds, halving the interval at each step, until the bracket has width 0.1 (for a 1 d.p. answer).

Test the midpoint at one extra decimal place. Once the bracket is 0.1 wide — say between 2.5 and 2.6 — test the midpoint x = 2.55. This is the boundary between the two possible rounded values and tells you definitively which one is correct.

State your conclusion. Write the answer to the required accuracy and explain in a sentence why you chose that value.

How do you solve x³ + x = 20 to 1 decimal place?

This is a classic KS3 trial and improvement problem. Let f(x) = x³ + x; you want the value of x where f(x) = 20.

Step 1 — find a bracket using integers:

Try x = 2: f(2) = 2³ + 2 = 8 + 2 = 10. Too small — result is below 20.

Try x = 3: f(3) = 3³ + 3 = 27 + 3 = 30. Too big — result is above 20.

The solution lies between x = 2 and x = 3.

Step 2 — test the midpoint x = 2.5:

f(2.5) = 2.5³ + 2.5 = 15.625 + 2.5 = 18.125. Too small.

The solution lies between x = 2.5 and x = 3.

Step 3 — test x = 2.6:

f(2.6) = 2.6³ + 2.6 = 17.576 + 2.6 = 20.176. Too big.

The solution lies between x = 2.5 and x = 2.6.

Step 4 — test the midpoint x = 2.55:

f(2.55) = 2.55³ + 2.55 = 16.581 + 2.55 = 19.131. Too small.

The solution lies between x = 2.55 and x = 2.6.

Step 5 — conclude: Every value in the interval (2.55, 2.6) rounds to 2.6 when expressed to 1 decimal place. Since f(2.55) < 20 and f(2.6) > 20, the solution must lie in that interval.

Answer: x ≈ 2.6 (to 1 d.p.)

How do you record your trials in a table?

Using a table is the clearest and most mark-friendly way to present trial and improvement. A well-laid-out table shows every trial, the computed result, and whether that result was too big or too small.

Here is the completed table for x³ + x = 20:

x | x³ + x | Compared to 20
2 | 10 | Too small
3 | 30 | Too big
2.5 | 18.125 | Too small
2.6 | 20.176 | Too big
2.55 | 19.131 | Too small
Conclusion | Solution in (2.55, 2.6) | x ≈ 2.6 to 1 d.p.

Never skip the "Compared to 20" column, even when the direction of adjustment feels obvious — it shows the examiner your reasoning and earns method marks independently of whether your arithmetic is perfect. Always end with a brief concluding sentence: "Since f(2.55) < 20 < f(2.6), the solution lies between 2.55 and 2.6, so x = 2.6 correct to 1 decimal place."

Why is the midpoint test the crucial final step?

The midpoint test is the step that resolves the rounding decision, and it is the one students most commonly miss. Many students narrow the bracket to width 0.1 and then stop — but that is insufficient.

Suppose the solution lies somewhere between 2.5 and 2.6. Without the midpoint test, you cannot know whether the answer rounds to 2.5 or 2.6. It could be 2.51 (which rounds to 2.5) or 2.57 (which rounds to 2.6). The midpoint 2.55 is the precise boundary between those two values.

By testing x = 2.55:

If f(2.55) is too small, the solution lies in (2.55, 2.6) — every value in that interval rounds to 2.6.

If f(2.55) is too big, the solution lies in (2.5, 2.55) — every value in that interval rounds to 2.5.

Testing the midpoint settles the rounding question definitively. It must always be the last trial you perform, and your concluding sentence must reference its result.

What mistakes should you avoid in trial and improvement?

Mistake | Why it matters | How to fix it
Skipping the midpoint test | The rounding is ambiguous — answer may be wrong | Always test the midpoint of the final 0.1 bracket as the last step
Not showing all trials | Loses method marks even if the final answer is correct | Record every trial in a full table with the "too big / too small" column
Rounding intermediate values | Leads to slightly wrong results and possible incorrect conclusions | Keep the full calculator display throughout; only round the final answer
Adjusting in the wrong direction | Wastes trials and can misplace the bracket | Too small → increase x; too big → decrease x
No concluding sentence | Loses the explanation mark | Write "x = 2.6 to 1 d.p. because f(2.55) < 20 < f(2.6)"

Frequently asked questions

How do I choose a good starting value for trial and improvement?

Start by trying small integers — 1, 2, 3, and 4. Substitute each into the equation and check whether the result is below or above the target. Once you find one integer giving a result that is too small and another giving a result that is too big, you have your initial bracket. If the equation involves x³, remember that 1³ = 1, 2³ = 8, 3³ = 27, and 4³ = 64, so you can often estimate the bracket quickly without a calculator. Some questions provide a starting hint — always use it.

Why must I test the midpoint at the end?

The midpoint test resolves the ambiguity between two adjacent 1 d.p. values. When the bracket is (2.5, 2.6), the midpoint is 2.55 — the exact boundary at which rounding switches from 2.5 to 2.6. Testing 2.55 tells you which half of the bracket holds the solution: if f(2.55) is too small, the solution is in (2.55, 2.6) and rounds to 2.6; if too big, it is in (2.5, 2.55) and rounds to 2.5. Without this test, you cannot state the correct rounded answer with certainty.

Can trial and improvement give an exact answer?

Generally, no. Trial and improvement gives an approximate answer to a stated degree of accuracy. For most KS3 equations of the form x³ + bx = c, the solution is irrational and cannot be expressed exactly as a fraction or terminating decimal. The method closes in on the true value without ever reaching it exactly. The answer x ≈ 2.6 to 1 d.p. is as precise as the method requires — the true value lies somewhere in (2.55, 2.6), and that is sufficient for KS3.

Is trial and improvement still used at GCSE?

Trial and improvement is primarily a KS3 technique. At GCSE, iterative numerical methods become more formal: you are typically given a rearranged equation in the form xₙ₊₁ = g(xₙ) and asked to apply it repeatedly, starting from a given seed value, until the answer converges to a required accuracy. The underlying logic is identical to trial and improvement — substitute, evaluate, repeat — so a thorough grounding at KS3 makes the GCSE version straightforward.

For Socratic algebra practice at KS3, see aitutors.me.