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Trial and Improvement

Learn how to use the trial and improvement method to solve equations by choosing values for variables and improving them iteratively.

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Trial and Improvement

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  1. Trial and Improvement 23 October 2019

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

  3. Finding the width of a rectangle x Area = 82 cm2 x + 1 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

  4. A more usual exam type question follows: Copy the table and you MUST show your trials. Example x2 + x = 240

  5. 1. x2 + x = 702 2. x2 – x = 420

  6. 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.

  7. 1. x2 + x = 66 2. x2 – x = 133

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