How do you find a one-decimal place approximation for #-root3 35#?

1 Answer
Oct 12, 2015

Answer:

Consider #(3+t)^3# to find #-root(3)(35)~~-3.3#

Explanation:

We know that #3^3=27#, so consider #(3+t)^3#...

#(3+t)^3=3^3+3*3^2t+3*3t^2+t^3#

#=27+27t+9t^2+t^3#

#~~27+27t# for small values of #t#

So approximating...

#35=27+27t#

Hence

#t=(35-27)/27=8/27~~0.3#

So #root(3)(35)~~3+0.3=3.3# and #-root(3)(35)~~-3.3#

More generally, to find the cube root of a number #n#, starting with an approximation #a_0#, iterate using the formula:

#a_(i+1) = a_i + (n - a_i^3)/(3a_i^2)#

This is a form of Newton Raphson method. Given a function #f(x)# for which we want to solve #f(x) = 0#, start with an approximation #a_0# and iterate using:

#a_(i+1) = a_i - f(a_i)/(f'(a_i))#

In our case #f(x) = x^3 - n# and #f'(x) = 3x^2#