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2. Intro Trial and Improvement Some non-linear equations, (like1, 2 and 3) below, may have integer solutions. Most equations however (like 4 and 5) do not have integer solutions and where this is the case we often have to be satisfied with an approximate solution. x 2 + x = 6 x 3 + 2x = 0 x = 2 or - 3 x = 0 2x 2 + x = 28x = - 4 Non-linear EquationsInteger Solutions Non-integer Solutions x 2 + x = 5 x 3 - 2x = 8 x =1.7913…… x = 2.3307..…. Non-linear Equations 1.8 or 1.79 seems a reasonable approximation 2.3 or 2.33 seems a reasonable approximation The method used in finding approximate solutions to these equations is called trial and improvement. We will concern ourselves only with finding a single positive solution in each case. 1 2 3 4 5

3. 1 d.p. Trial and Improvement The Method Solve x 2 - x = 27 (to 1 d.p) x x 2 - x 27 55 2 - 5 = 20 too small 66 2 - 6 = 30 too big 1.Make a table similar to this one. 27 is your target number. 2. Make an intelligent guess to find two positive consecutive integers that output values that straddle the target number. 3. Repeat above with consecutive 1 d.p. numbers between 5 and 6. (Trying 5.5 first) 5.55.5 2 - 5.5 = 24.75 too small 5.65.6 2 - 5.6 = 25.76 too small 5.75.7 2 - 5.7 = 26.79 too small 5.85.8 2 - 5.8 = 27.84 too big 4. One of these is the correct 1 d.p solution but which one? 5. Compute the mid-point value to help you decide. 5.75 2 - 5.75 = 27.3125 5.75 too big Worked Example 1. 5.7 5.8 5.75 So the true value must lie in here and all values are 5.7 when rounded. x = 5.7

4. Trial and Improvement The Method Solve x 3 + 2 x = 6 (to 1 d.p) x x 3 + 2 x 6 11 3 + 2 x 1 = 3too small 22 3 + 2 x 2 = 12 too big 1.Make a table similar to this one. 6 is your target number. 2. Make an intelligent guess to find two positive consecutive integers that output values that straddle the target number. 3. Repeat above with consecutive 1 d.p. numbers between 1 and 2.(Trying 1.3) 1.31.3 3 + 2 x 1.3 = 4.797 too small 1.41.4 3 + 2 x 1.4 = 5.544too small 5. Compute the midpoint output. 1.5 too big x = 1.5 Worked Example 2. 1.5 3 + 2 x 1.5 = 6.375 1.45 too small 1.45 3 + 2 x 1.45 = 5.948… 1.4 1.5 1.45 So the true value must lie in here and all values are 1.5 when rounded. Midpoint value: too big  round down too small  round up 4. One of these is the correct 1 d.p solution but which one?

5. Trial and Improvement The Method x x 2 + 3 x 52 55 2 + 3 x 5 = 40too small 66 2 + 3 x 6 = 54 too big 1.Make a table similar to this one. 52 is your target number. 2. Make an intelligent guess to find two positive consecutive integers that output values that straddle the target number. 3. Repeat above with consecutive 1 d.p. numbers between 5 and 6.(Trying 5.9) 5.95.9 2 + 3 x 5.9 = 52.51 too big 5.85.8 2 + 3 x 5.8 = 51.04too small 5. Compute the midpoint output. 5.85 too small x = 5.9 Worked Example 3. 5.85 2 + 3 x 5.85 = 51.7725 6. Too small so round up x x + 3 52 cm 2 Find the width of the rectangle to 1 d.p. 4. One of these is the correct 1 d.p solution. Remember :midpoint value: too big  round down too small  round up

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

7. 2 d.p. Trial and Improvement x x 2 + 3 x 52 55 2 + 3 x 5 = 40too small 66 2 + 3 x 6 = 54 too big 5.95.9 2 + 3 x 5.9 = 52.51 too big 5.85.8 2 + 3 x 5.8 = 51.04too small x = 5.87 Worked Example Solve x 2 + 3 x = 52 (to 2 d.p) In solving problems to 2 d.p. we simply continue the process one stage further. The problem shown has a 1d.p. solution between 5.8 and 5.9. Two d.p. problem 1. We now need to find 2 consecutive 2 d.p. numbers that straddle the target number. Trying 5.86 gives: 5.865.86 2 + 3 x 5.86 = 51.9196too small 5.875.87 2 + 3 x 5.87 = 52.0669too big 3. Calculate the midpoint output. 5.8655.865 2 + 3 x 5.865 = 51.9…too small 4. Too small so round up. 2. One of these is the correct 2 d.p solution. Remember :midpoint value: too big  round down too small  round up

8. Questions 2 Trial and Improvement 1. x 2 + x = 15 Questions Solve the following problems below by finding a positive solution for x to 2 d.p. x = 3.41 2. x 3 - x = 100 x = 4.71 3. 3 x 2 + 2 x = 92 x = 5.21 x 8 cm 2 x + 5 x = 1.27 4. x 9 cm 2 x = 2.39 5. 2x - 1

9. Worksheet 1. x 2 + x = 8 2. x 3 - x = 180 3. 2 x 2 - 3 x = 1 x 27 cm 2 x + 4 4. x 11 cm 2 5. 2x + 1 Questions 1 d.p. 1. x 2 + x = 15 2. x 3 - x = 100 3. 3 x 2 + 2 x = 92 x 8 cm 2 x + 5 4. x 9 cm 2 5. 2x - 1 Questions 2 d.p. worksheet