GCSE Maths Starter 15 20−12÷4 Solve 4x + 3 = 18 – 2x Tom, Tim and Bill share £60 in the ratio 2:3:5. Calculate the amount each person receives. p = mq + r, make q the subject Find the mean, median and mode and range of: 5, 11, 5, 7, 9, 10, 4, 4, 5, 10

Lesson 15 Trial and improvement Mathswatch clip (110). To estimate a solution to an equation (Grade D/C ) EXTN: To form and solve equations using trial and improvement(Grade C)

1 d.p. Solve x2 - x = 27 (to 1 d.p) x x2 - x 27 x = 5.7 Worked Example 1. Trial and Improvement The Method Solve x2 - x = 27 (to 1 d.p) Make a table similar to this one. 27 is your target number. x x2 - x 27 2. Make an intelligent guess to find two positive consecutive integers that output values that straddle the target number. 5 52 - 5 = 20 too small 6 62 - 6 = 30 too big 5.5 5.52 - 5.5 = 24.75 too small 5.6 5.62 - 5.6 = 25.76 too small 3. Repeat above with consecutive 1 d.p. numbers between 5 and 6. (Trying 5.5 first) 5.7 5.72 - 5.7 = 26.79 too small 5.8 5.82 - 5.8 = 27.84 too big 4. One of these is the correct 1 d.p solution but which one? too big 5.75 5.752 - 5.75 = 27.3125 So the true value must lie in here and all values are 5.7 when rounded. x = 5.7 5. Compute the mid-point value to help you decide. 1 d.p. 5.7 5.8 5.75

Remember :midpoint value: too big  round down too small  round up Worked Example 3. Trial and Improvement The Method x + 3 Make a table similar to this one. 52 is your target number. Find the width of the rectangle to 1 d.p. x 52 cm2 2. Make an intelligent guess to find two positive consecutive integers that output values that straddle the target number. x x2 + 3x 52 5 52 + 3 x 5 = 40 too small 6 62 + 3 x 6 = 54 too big 5.9 5.92 + 3 x 5.9 = 52.51 too big 3. Repeat above with consecutive 1 d.p. numbers between 5 and 6.(Trying 5.9) 5.8 5.82 + 3 x 5.8 = 51.04 too small 5.85 5.852 + 3 x 5.85 = 51.7725 too small 4. One of these is the correct 1 d.p solution. x = 5.9 5. Compute the midpoint output. Remember :midpoint value: too big  round down too small  round up 6. Too small so round up

Questions 1 x = 2.4 x = 5.7 x = 1.8 x = 3.6 x = 2.1 1. x2 + x = 8 Trial and Improvement Questions Solve the following problems below by finding a positive solution for x to 1 d.p. 1. x2 + x = 8 x = 2.4 2. x3 - x = 180 x = 5.7 3. 2x2 - 3x = 1 x = 1.8 x + 4 4. x = 3.6 x 27 cm2 2x + 1 Questions 1 x 11 cm2 x = 2.1 5.

has a solution between 4 and 5 Foundation/Higher Exam Type Question Calculator): Leave blank Q1 (Total 4 marks) 1. (4 marks) x = ......................... The equation x3 - 12 = 85 has a solution between 4 and 5 Use a trial and improvement method to find the solution, correct to one decimal place. You must show all of your working.

3) width of rectangle is 7.8cm 4) x = 3.7 N 29 Trial & Improvement Answers Exercise 1 1) Generate the equation x(x + 5) = 73 from the information given (length times width of rectangle gives its area). Use trial & improvement to arrive at a solution of 6.4cm 2) x = 5.3 3) width of rectangle is 7.8cm 4) x = 3.7 Exam Q1 4.6 Exam Q2 4.7 Exam Q3 5.8 (note: This is closer than 5.9)

Lesson 15 Trial and improvement Mathswatch clip (110). Exam questions