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 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 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 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. 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 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 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 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 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 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 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 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 = 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 = 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 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 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 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?
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