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

Oct 23, 2015

Find $2 < \sqrt[3]{15} < 3$, 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 $2 < \sqrt[3]{15} < 3$

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:

${a}_{i + 1} = {a}_{i} + \frac{n - {a}_{i}^{3}}{3 {a}_{i}^{2}}$

In our case $n = 15$ and choose ${a}_{0} = 2.5$ since the cube root is somewhere between $2$ and $3$.

Then:

${a}_{1} = {a}_{0} + \frac{n - {a}_{0}^{3}}{3 {a}_{0}^{2}} = 2.5 + \frac{15 - {2.5}^{3}}{3 \cdot {2.5}^{2}}$

$= 2.5 + \frac{15 - 15.625}{18.75} = 2.5 - \frac{0.625}{18.75} = 2.4 \dot{6} \dot{6}$

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