# Mathmatics

posted by
**Flamingo**
.

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.