Trial and Improvement 13 April 2019

x – 1 Volume = 344 cm3 x + 4 x …………………………………. WIDTH (x) = …………….cm

Finding the width of a rectangle In the following rectangle the length is 1 cm more than the width. Its area is 82 cm2. Let the width be x. Choose a starting value for x If x = 8 Area = 8 x 9 = 72 cm2 If x = 9 Area = 9 x 10 = 90 cm2 Choose a value to 1 d.p. between 8 and 9. What decimal value of x is closest? x = 8.6 cm x + 1 x Area = 82 cm2

A more usual exam type question follows: Copy the table and you MUST show your trials. Example x2 + x = 240 Trial x value x2 + x COMMENT 12 122 + 12 = 156 Too small 16 162 + 16 = ___

1. x2 + x = 702 x value x2 + x COMMENT 2. x2 – x = 420 x value x2 – x COMMENT

Trial and Improvement This method of solving an equation is called trial and improvement. Its main features are: Choose a whole number of x Improve the choice by noting how far the answer is from what is needed. If the answer is a decimal, find the value x to 1 decimal place.

1. x2 + x = 66 x value x2 + x COMMENT 2. x2 – x = 133 x value x2 – x COMMENT