hi
I'm confused with this question
it says
Using a trial and improvement method, slove the following equataions giving your answers to 1 decimal place
a) a squared - 28 = 0
b) a squared - 104= 0
c) a cubed - a = 15
how ?
start with some estimation
28 is a bit larger than 5^2, which would have been 5^2
so try 5.1^2 which is 26.01, too small
try 5.2^2 to get 27.04, still too small
5.3^2 = 28.09, almost dead on.
so to one decimal place a = 5.3
do the same with the other questions.
Using a trial and improvement method, slove the following equataions giving your answers to 1 decimal place
a) a squared - 28 = 0
b) a squared - 104= 0
c) a cubed - a = 15
n.........n^2
5.........25
a.........28
6.........36
(6 - a)/(6 - 5) = (36 - 28)/(36 - 25)
6 - a = 8/11
66 - 11a = 8 making a = 5.272
5.272^2 = 27.79
n.......n^2
5.272..27.79
a.......28
6.......36
(6 - a)/(6 - 5.272) = (36 - 28)/(36 - 27.79) making a = 5.2906
5.29062 = 27.99
Carry it as far as you wish.
Try the others now that you have a method.
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To solve the equations given using a trial and improvement method, you can follow these steps:
a) Equation: a^2 - 28 = 0
- Start with some estimation by recognizing that 28 is a bit larger than 5^2, which is 25.
- Try a value slightly higher than 5, such as 5.1. Calculate 5.1^2 = 26.01, which is still too small.
- Increase the value slightly again and calculate 5.2^2 = 27.04, which is still too small.
- Continue this process until you reach a value that is very close to 28. Calculating 5.3^2 = 28.09, which is almost dead on.
- Therefore, to one decimal place, a = 5.3.
b) Equation: a^2 - 104 = 0
- Follow the same process as above.
- Estimate using 10^2 = 100, and since 104 is slightly larger, start with value greater than 10.
- Try 10.1^2 = 102.01, which is still too small.
- Increase the value slightly again and calculate 10.2^2 = 104.04, which is close.
- Continuing this process, you will find that to one decimal place, a = 10.2.
c) Equation: a^3 - a = 15
- This is a cubic equation, so the process will be slightly different.
- Start with an estimation and try the value of a = 5.
- Calculate 5^3 - 5 = 125 - 5 = 120, which is larger than 15.
- Increase the value slightly and try a = 6.
- Calculate 6^3 - 6 = 216 - 6 = 210, which is still too large.
- Use the estimation values to set up a proportion:
(6 - a) / (6 - 5) = (36 - 15) / (36 - 125)
- Solve the proportion to find a = 5.272.
- Continue this process if you want a higher level of accuracy, carrying it as far as you wish.
By following the trial and improvement method, you can solve equations by estimating values, testing them, and refining your estimation until you find a solution. Apply this method to the given equations and adjust the values until you find the solutions to one decimal place.