# How do you find a one-decimal place approximation for #sqrt 18#?

##### 1 Answer

#### Answer:

Use Newton Raphson method to find

#### Explanation:

To find an approximation for the square root of a number

#a_(i+1) = (a_i^2+n)/(2a_i)#

In our case

Then:

#a_1 = (a_0^2+n)/(2a_0) = (4^2+18)/(2*4) = (16+18)/8 = 34/8 = 17/4 = 4.25#

Unfortunately

#a_2 = (a_1^2+n)/(2a_1) = ((17/4)^2+18)/(2*(17/4))#

#=(289/16+18)/(17/2) =(289+18*16)/(17*8)=(289+288)/136#

#=577/136 ~~ 4.24265#

That's more decimal places than we need, but at least we can say with confidence that

Alternatively, you can use the method described in http://socratic.org/questions/given-an-integer-n-is-there-an-efficient-way-to-find-integers-p-q-such-that-abs-

to find a continued fraction expansion for

#sqrt(18) = [4;bar(4,8)] = 4+1/(4+1/(8+1/(4+1/(8+1/(4+...)))))#

Then you can approximate

For example:

#sqrt(18) ~~ [4;4,8] = 4+1/(4+1/8) = 4+8/33 = 140/33 = 4.dot(2)dot(4)#