Trial and Improvement We use the trial and

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Trial and Improvement We use the “trial and improvement” method to solve equations. Examples

Trial and Improvement We use the “trial and improvement” method to solve equations. Examples Solve the following equations using the trial and improvement method Question 1 Question 2 Question 3

1. 2 y + 8 = 20 The aim is to find the value

1. 2 y + 8 = 20 The aim is to find the value of y. By trial and improvement……… Let’s try y=5 (2 x 5) + 8 = 18 Let’s try y=7 (2 x 7) + 8 = 22 Let’s try Too small Too big y=6 (2 x 6) + 8 = 20 Got it! So the solution of the equation is y=6

2. 4 x - 6 = 30 Our aim is to find the value

2. 4 x - 6 = 30 Our aim is to find the value of x. By trial and improvement……… Let’s try x = 10 (4 x 10) - 6 = 34 Let’s try x=8 (4 x 8) - 6 = 26 Let’s try Too big Too small x=9 (4 x 9) - 6 = 30 Got it! So the solution of the equation is x=9

3. d + 6=8 2 Our aim is to find the value of d.

3. d + 6=8 2 Our aim is to find the value of d. By trial and improvement……… Let’s try d=6 (6 ÷ 2) + 6 = 9 Let’s try d=2 (2 ÷ 2) + 6 = 7 Let’s try Too big Too small d=4 (4 ÷ 2) + 6 = 8 Got it! So the solution of the equation is d=4

Sometimes, solutions to equations are not whole numbers. Examples Solve the following equations, correct

Sometimes, solutions to equations are not whole numbers. Examples Solve the following equations, correct to 1 decimal place, by using the trial and improvement method. Question 1 Question 2 Question 3 Question 4 Exercises

1. x² + 5 = 25 Let’s try x=4 4²+ 5 = 21 Let’s

1. x² + 5 = 25 Let’s try x=4 4²+ 5 = 21 Let’s try Too small x=5 5²+ 5 = 30 Too big The value of x lies between 4 and 5. Let’s try x = 4· 5²+ 5 = 25· 25 Let’s try Too big x = 4· 4²+ 5 = 24· 36 Too small

The value of x lies between 4· 4 and 4· 5. Both values are

The value of x lies between 4· 4 and 4· 5. Both values are to one decimal place. So which x is our solution? To answer this question, one more step needs to be considered……to try x = 4· 45. Let’s try x = 4· 45²+ 5 = 24· 8025 Too small So the solution of the equation is x = 4· 5 (1 dp)

2. x² - 10 = 80 Let’s try x=9 9²- 10 = 71 Let’s

2. x² - 10 = 80 Let’s try x=9 9²- 10 = 71 Let’s try Too small x = 10 10²- 10 = 90 Too big The value of x lies between 9 and 10. Let’s try x = 9· 5²-10 = 80· 25 Let’s try Too big x = 9· 4²- 10 = 78· 36 Too small

The value of x lies between 9· 4 and 9· 5. Both values are

The value of x lies between 9· 4 and 9· 5. Both values are to one decimal place. So which x is our solution? To answer this question, one more step needs to be considered……to try x = 9· 45. Let’s try x = 9· 45²- 10 = 79· 3025 Too small So the solution of the equation is x = 9· 5 (1 dp)

3. 5 a²= 70 Let’s try a=2 5 x 2² = 20 Let’s try

3. 5 a²= 70 Let’s try a=2 5 x 2² = 20 Let’s try a=4 5 x 4²= 80 Let’s try Too small Too big a=3 5 x 3²= 45 Too small The value of a lies between 3 and 4. Let’s try a = 3· 5 5 x 3· 5²= 61· 25 Too small

Let’s try a = 3· 7 5 x 3· 7²= 68· 45 Let’s try

Let’s try a = 3· 7 5 x 3· 7²= 68· 45 Let’s try Too small a = 3· 8 5 x 3· 8²= 72· 2 Too big The value of a lies between 3· 7 and 3· 8. Let’s try x = 3· 75 5 x 3· 75² = 70· 3125 Too big So the solution of the equation is a = 3· 7 (1 dp)

4. 3 a²- a = 30 Let’s try a=2 (3 x 2²) - 2

4. 3 a²- a = 30 Let’s try a=2 (3 x 2²) - 2 = 10 Let’s try a=3 (3 x 3²) - 3 = 24 Let’s try Too small a=4 (3 x 4²) - 4 = 44 Too big The value of a lies between 3 and 4. Let’s try a = 3· 5 (3 x 3· 5²) - 3· 5 = 33· 25 Too big

Let’s try a = 3· 4 (3 x 3· 4²) - 3· 4 =

Let’s try a = 3· 4 (3 x 3· 4²) - 3· 4 = 31· 28 Let’s try Too big a = 3· 3 (3 x 3· 3²) – 3· 3 = 29· 37 Too small The value of a lies between 3· 3 and 3· 4. Let’s try a = 3· 35 (3 x 3· 35²) – 3· 35 = 30· 3175 Too big So the solution of the equation is a = 3· 3 (1 dp)

Solve the following equations, correct to 1 decimal place, by using the trial and

Solve the following equations, correct to 1 decimal place, by using the trial and improvement method. 1. x² + 9 = 20 2. a² - 24 = 100 3. 8 x² = 220 4. 6 a² + a = 110

Problems involving Trial and Improvement Examples 1. The area of a rectangle is 22

Problems involving Trial and Improvement Examples 1. The area of a rectangle is 22 cm². The width of the rectangle is y cm long. The length of the rectangle is 3 cm more than the width. Find the width of the rectangle, correct to 1 decimal place, by using the trial and improvement method.

2. A square has an area of 32 cm². Use a trial and improvement

2. A square has an area of 32 cm². Use a trial and improvement method to calculate the length of one side. 3. The perpendicular height of a right angled triangle is 5 cm more than its base. If its area is 92 cm², what is the length of its base?