Trial and Improvement Objectives: C Grade Form and solve equations such as x2 + x = 12 using trial and improvement Prior knowledge: Rounding to decimal places Substitution into algebraic expressions The shape of quadratic / cubic graphs Use of the bracket button on a calculator

Trial and Improvement Estimate the square root of the following numbers: 17 30 47 68 110 83 4.1 4.123105626 5.5 5.477225575 6.9 6.8556546 8.2 8.246211251 10.5 10.48808848 9.1 9.110433579 Now check your answers on a calculator

Find the positive solution to the equation x2 - x = 60 y = x2 - x Trial and Improvement Find the positive solution to the equation x2 - x = 60 give your answer to 1 decimal place If we consider this drawing a graph we know the solution can be found by drawing the line y = 60 and y = x2 – x, finding the value of x at the point of intersection. y = 60 because of the scale of the graph we have to use we cannot find the value of x to 1 d.p. but we can see it is between 8 and 9.

Trial and improvement is where we try values of x in the equation and try to get as close to the given value for y as possible x2 – x = 60 Try x = 8 Now we know that the solution is between 8.2 and 8.3 we try 8.25 64 – 8 = 56 x = 8 is too low Try x = 9 81 – 9 = 72 x = 9 is too low Try x = 8.5 72.25 – 8.5 = 63.75 x = 8.5 is too high Try x = 8.3 Now even the expanding graph is not big enough for the level of accuracy required 68.89 – 8.3 = 60.59 x = 8.3 is too high Try x = 8.2 67.42 – 8.2 = 59.04 x = 8.2 is too low

Trial and improvement is where we try values of x in the equation and try to get as close to the given value for y as possible Now we know that the solution is between 8.2 and 8.3 we try 8.25 Try x = 8.25 68.0625 – 8.25 = 59.8125 x = 8.25 is too low We need the value of x to 1 d.p. We know the solution is now between 8.25 and 8.3. Try x = 8.3 Now even the expanding graph is not big enough for the level of accuracy required Any value between 8.25 and 8.3 would be rounded to 8.3 to 1 d.p 68.89 – 8.3 = 60.59 x = 8.3 is too high Try x = 8.2 Therefore to 1 d.p x = 8.3 67.42 – 8.2 = 63.75 x = 8.2 is too low

Find the value of x to 1.d.p to solve this equation: x3 + x = 12 Trial and Improvement Find the value of x to 1.d.p to solve this equation: x3 + x = 12 We can now do this as a table: Trial Value of x x3 x x3-x Comment   3 27 3 30 Too High 2 8 2 10 Too Low 2.1 9.261 2.1 11.36 Too Low 2.2 10.65 2.2 12.85 Too High 2.15 9.938 2.15 12.09 Too High 2.14 9.8 2.14 11.94 Too Low Because 2.15 is too high and any number less than 2.15 would be rounded to 2.1 to 1 d.p. Finding the answer when x = 2.14 proves that this is correct.

Trial and Improvement Now do these: Find the positive solutions to 1 decimal place 1. x2 + 2x = 63 2. x2 - 2x = 675 3. x3 + 2x = 520 4. x5 + x = 33 768 5. x2 - 7x = 368 6. x - x3 = -336

Trial and Improvement Worksheet x2 + 2x = 63 x5 + x = 33 768 Trial Value of x x2 2x x2 + 2x Comment x5 x x5 + x   x2 - 2x = 675 x2 - 7x = 368 x2 - 2x 7x x2 - 7x x3 + 2x = 520 x - x3 = -336 x3 x3 + 2x x - x3

Trial and Improvement x = 7 x = 8.0 x = 27 x = 23 x = 8.0 x = 7.3 63 x5 + x = 33 768 33768 Trial Value of x x2 2x x2 + 2x Comment x5 x x5 + x 5 25 10 35 Too Low 8 32768 16 32784 6 36 12 48 9 59049 18 59067 Too High 7 49 14   8.1 34868 16.2 34884 8.05 33805 16.1 33821 8.02 33180 16.04 33196 8.04 33595 16.08 33612 x2 - 2x = 675 675 x2 - 7x = 368 368 x2 - 2x 7x x2 - 7x 20 400 40 360 625 175 450 30 900 60 840 22 484 154 330 28 784 56 728 23 529 161 26 676 52 624 27 729 54 x3 + 2x = 520 520 x - x3 = -336 -336 x3 x3 + 2x x - x3 512 528 343 -294 357 64 -448 7.5 421.9 15 436.9 56.25 -366 7.9 493 15.8 508.8 7.4 54.76 405.2 -350 7.95 502.5 15.9 518.4 7.3 53.29 389 7.96 504.4 15.92 520.3 x = 8.0 x = 27 x = 23 x = 8.0 x = 7.3