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Trial and Improvement

Trial and Improvement. Objectives: C Grade Form and solve equations such as x 2 + x = 12 using trial and improvement. Prior knowledge: Rounding to decimal places Substitution into algebraic expressions The shape of quadratic / cubic graphs

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Trial and Improvement

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  1. 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

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

  3. Trial and Improvement Find the positive solution to the equation x2 - x = 60 give your answer to 1 decimal place y = x2 - x 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.

  4. Trial and Improvement 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 Now we know that the solution is between 8.2 and 8.3 we try 8.25 Try x = 8 x = 8 is too low 64 – 8= 56 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 x = 8.2 is too low 67.42 – 8.2= 59.04

  5. Trial and Improvement 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 x = 8.25 is too low 68.0625 – 8.25= 59.8125 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 x = 8.2 is too low 67.42 – 8.2= 63.75

  6. 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: 27 3 3 30 Too High 8 2 2 10 Too Low 2.1 9.261 2.1 11.36 Too Low 12.85 10.65 2.2 2.2 Too High 9.938 2.15 12.09 2.15 Too High 9.8 2.14 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.

  7. 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

  8. Trial and Improvement Worksheet

  9. Trial and Improvement x = 7 x = 8.0 x = 27 x = 23 x = 8.0 x = 7.3

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