Module 6 (M6) – Algebra - Trial and improvement

Before reading this guide, it may be helpful to read the guides from Module 1 (M1) on decimals and M2 Rounding to significant figures.

Trial and improvement

Trial and improvement is a method for solving equations when there is no exact answer.

You may also be asked to give the solution to a given number of decimal places or significant figures. The question will usually indicate the degree of accuracy required.

When solving an equation by trial and improvement:

Choose a sensible 'starting value'.
Substitute this value into the left hand side (LHS) of the equation
Use a calculator to evaluate it
Decide whether the value obtained is too big or too small
Choose an improved value and re-evaluate
Repeat steps 2–5 until the answer and the previous one round to the given accuracy

This process is usually recorded on a table.

Example

Using trial and improvement, solve the equation \(x^{3} + x = 20\)

Give your answer correct to one decimal place.

Solution

First, choose a starting value.

\(x = 3\) is clearly too big as \(x^{3}\) is 27 and the target value is 20.
\(x = 2\) is a sensible starting value

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Image caption,
The result is too small so choose a larger value. Try 2.5
Image caption,
The result is getting closer but is still too small. Try 2.6
Image caption,
The result is now too big. X must lie between 2.5 and 2.6. Try 2.55, which is halfway between 2.5 and 2.6.
Image caption,
2.6 is too big and 2.55 is too small so the answer must lie in between these two values.

Any value inbetween 2.55 and 2.6 will round to 2.6.

Answer

\(x = 2.6\) (correct to one decimal place).

To get all of the marks, you must show all the steps.

Question

The table below shows the start of a trial and improvement method to solve the equation \(2x^{3} – x = 40\).

Continue the process and give your answer to one decimal place.

An image of a table showing trial and improvement., with a header row which reads: "trial value of x; value of 2x cubed – x, target is 40; Result. First row reads 3; 2(3 cubed) – 3 = 51; too big." Second row, only the first column is completed with 2.

Question

Using trial and improvement, solve the equation \(x^{4} – x – 140 = 0\).

Give your answer correct to two decimal places.

Question

Use trial and improvement to find a solution of the equation \(x^{3} + 2x = 22\).

Give your answer correct to 2 decimal places. You must show all your working.

Test yourself

Question 1

Use trial and improvement to find a solution to the equation \(x^2 + \frac {x}{2} = 17\)Give your answer correct to 1 decimal place.

You should show all your working in exam conditions

Solution:

Trial value of x	Value of \(x^2 + \frac{x}{2}\) Target is 17	Result
3	\(3^2 + \frac{3}{2} = 10.5\)	Too small
4	\(4^2 + \frac{4}{2} = 18\)	Too big
3.5	\(3.5^2 + \frac{3.5}{2} = 14\)	Too small
3.6	\(3.6^2 + \frac{3.6}{2} = 14.76\)	Too small
3.7	\(3.7^2 + \frac{3.6}{2} = 15.54\)	Too small
3.8	\(3.8^2 + \frac{3.8}{2} = 16.34\)	Too small
3.9	\(3.9^2 + \frac{3.9}{2} = 17.16\)	Too big
3.85	\(3.85^2 + \frac{3.85}{2} = 16.7475\)	Too small

The solution is between 3.85 and 3.9. Any number in this range rounds to 3.9

Answer x = 3.9 to 1 d.p.

Question 2

Use trial and improvement to find a solution to the equation \(\frac{x^2}{2}\ +\ x\ = 35\)

Give your answer correct to 1 decimal place

You should show all your working in exam conditions

Solution:

Trial value of x	Value of \(\ \frac{x^2}{2}\ +\ x\) Target is 35	Result
8	\(\frac{8^2}{2}\ +\ 8\ =\ 40\)	Too big
7	\(\frac{7^2}{2}\ +\ 7\ =\ 31.5\)	Too small
7.5	\(\frac{{7.5}^2}{2}\ +\ 7.5\ =\ 35.625\)	Too big
7.4	\(\frac{{7.4}^2}{2}\ +\ 7.4\ =\ 34.78\)	Too small
7.45	\(\frac{{7.45}^2}{2}\ +\ 7.45\ =\ 35.20125\)	Too big

7.4 is too small and 7.45 is too big so the solution is somewhere in between.
Any number in this range will round to 7.4.

Answer x = 7.4 correct to 1 d.p.

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