How do you find a one-decimal place approximation for root3 15?

1 Answer
George C.
Oct 23, 2015

Find 2 < root(3)(15) < 3, then using 2.5 as a first approximation and one step of Newton's method, find that this is already good to one decimal place.

Explanation:

Note that 2^3 = 8 < 15 < 27 = 3^3

So 2 < root(3)(15) < 3

To find an approximation for the cube root of a number n, choose a reasonable first approximation a_0 and apply the following formula (from Newton's method), repeatedly if necessary:

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

In our case n = 15 and choose a_0 = 2.5 since the cube root is somewhere between 2 and 3.

Then:

a_1 = a_0 + (n - a_0^3)/(3a_0^2) = 2.5 + (15 - 2.5^3)/(3*2.5^2)

=2.5 + (15-15.625)/18.75 = 2.5 - 0.625/18.75 = 2.4dot(6)dot(6)

So our first approximation was already good to one decimal place.

Related questions
See all questions in Square Roots and Irrational Numbers
Impact of this question
3394 views around the world
You can reuse this answer
Creative Commons License