# Mathmatics

posted by .

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 ?

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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