Year 8 Mathematics Area and Perimeter http://www.mathsisfun.com/fractions.html Area and Perimeter
Learning Intentions Learning Intentions Understand what is meant by a trial Be able to calculate the value of expressions To select the next suitable value for a trial Be able to round to a number of decimal places Know when the solution has been reached
Solving Equations An equation such as 2x + 2 = 5 can be solved using algebra We subtract 2 from each side to give 2x = 3 And then divide each side by 2 to give x = 1.5
Trial and Improvement However, sometimes we are unable to solve equations using algebra In these cases we guess a solution to see if we are close to the answer. We use the result of our guess to help us get a better solution We continue this process until we have a solution to the accuracy we need This is called trial and improvement
Example Solve the equation x3 + 2x = 1 to 1 decimal place To do this we will guess a value for x, so let x = 1 If x = 1, x3 + 2x = 3 This is too big Let’s put this in a table
First Guess If x = 1 is too big, we need to try a smaller value x3 + 2x Comment 1 3 Too big If x = 1 is too big, we need to try a smaller value Let’s try x = 0
Second Guess Unfortunately this is too small x x3 + 2x Comment 1 3 Too big Too small Unfortunately this is too small But we now know that the answer lies between 0 and 1, so there is no point trying x values less than 0 or bigger than 1 Let’s try 0.5
In the Middle x x3 + 2x Comment 1 3 Too big Too small 0.5 1.125 We can continue this to try and find the values with 1 decimal point that x lies between
One Decimal Place x x3 + 2x Comment 1 3 Too big Too small 0.5 1.125 0.4 0.864 We now know that x lies between 0.5 and 0.5, but which is the answer closer to? It’s easy to guess 0.5, but this is not always the case. Let’s try 0.45
Two decimal places x x3 + 2x Comment 1 3 Too big Too small 0.5 1.125 0.4 0.864 0.45 0.991125 Since x = 0.45, gives a value that is too small we know that the answer lies between 0.45 and 0.5 Thus the solution to x3 + 2x = 1 is x = 0.5 (1 dp)