Join TSR
 
 About Us | FAQs | Sign in
 
HomeForumsArticlesRevision NotesPersonal StatementsUniversity GuidesHealth & RelationshipsPostgraduate StudyCareers
Join the UK's largest student community  
Advanced
Search

Join The Student Room Today

Be part of the UK's largest and fastest growing student community.

It's free to join and a lot of fun - Get inspired, express your ideas, interact and share

Join The Student Room Today

Revision:Solving Equations

From The Student Room

TSR Wiki > Study Help > Subjects and Revision > Revision Notes > Mathematics > Solving Equations

Contents

Trial and Improvement

Any equation can be solved by trial and improvement (/error). However, this is a tedious procedure.

Example: Solve t^3 + t = 17 by trial and improvement.

Firstly, select a value of t to try in the equation. I have selected t = 2. Put this value into the equation. We are trying to get the answer of 17.


If t = 2, t^3 + t = 2^3 + 2 = 10. This is lower than 17, so we try a higher value for t.

If t = 2.5, t^3 + t = 18.125 (too high)

If t = 2.4, t^3 + t = 16.224 (too low)

If t = 2.45, t^3 + t = 17.156 (too high)

If t = 2.44, t^3 + t = 16.966 (too low)

If t = 2.445, t^3 + t = 17.061 (too high)


So we know that t is between 2.44 and 2.445. So to 2 decimal places, t = 2.44.


Iteration

This is a way of solving equations. An iteration formula might look like the following:

\displaystyle x_{n+1} = 2 + \frac{1}{x_n}

You are usually given a starting value, which is called x_0.

If x_0 = 3, substitute 3 into the original equation where it says x_n. This will give you x_1. (This is because if n = 0, x_1 = 2 + \frac{1}{x_0} and x_0 = 3).

\displaystyle x+1 = 2 + \frac{1}{3} = 2.333333 (by substituting in 3).


To find x_2, substitute the value you found for x_1.

\displaystyle x_2 = 2 + \frac{1}{2.333 333} = 2.428 571


Repeat this until you get an answer to a suitable degree of accuracy. This may be about the 5th value for an answer correct to 3s.f. In this example, x_5 = 2.414...


Example

a} Show that \displaystyle x = 1 + \frac{11}{x - 3} is a rearrangement of the equation \displaystyle x^2 - 4x - 8 = 0.

b) Use the iterative formula \displaystyle x_{n+1} = 1 + \frac{11}{x_n - 3} together with a starting value of x_1 = -2 to obtain a root of the equation x^2 - 4x - 8 = 0 accurate to one decimal place.


Answers

a) multiply everything by (x - 3):

\displaystyle x(x - 3) = 1(x - 3) + 11

so \displaystyle x^2 - 3x = x + 8

so \displaystyle x^2 - 4x - 8 = 0


b) \displaystyle x_1 = -2


\displaystyle x_2 = 1 + \frac{11}{-2 - 3} (substitute -2 into the iteration formula)

\displaystyle = -1.2


\displaystyle x_3 = 1 + \frac{11}{-1.2 - 3} (substitute -1.2 into the above formula)

\displaystyle = -1.619


Then:

x_4 = -1.381

x_5 = -1.511

x_6 = -1.439

x_7 = -1.478

therefore, to one decimal place, x = 1.5.


Comments

This article is only concerned with looking at numerical methods - it could do with being renamed.

Another article is needed looking at general methods for solving (linear) equations. A link to the article on solving quadratic equations should be include as well as links to any other relevant pages.

Retrieved from "http://thestudentroom.co.uk/wiki/Revision:Solving_Equations"
Article Tools:  Create New Article  |  Report This Article  |  This Article's History  |