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

##### 1 Answer
Oct 12, 2015

Consider ${\left(3 + t\right)}^{3}$ to find $- \sqrt{35} \approx - 3.3$

#### Explanation:

We know that ${3}^{3} = 27$, so consider ${\left(3 + t\right)}^{3}$...

${\left(3 + t\right)}^{3} = {3}^{3} + 3 \cdot {3}^{2} t + 3 \cdot 3 {t}^{2} + {t}^{3}$

$= 27 + 27 t + 9 {t}^{2} + {t}^{3}$

$\approx 27 + 27 t$ for small values of $t$

So approximating...

$35 = 27 + 27 t$

Hence

$t = \frac{35 - 27}{27} = \frac{8}{27} \approx 0.3$

So $\sqrt{35} \approx 3 + 0.3 = 3.3$ and $- \sqrt{35} \approx - 3.3$

More generally, to find the cube root of a number $n$, starting with an approximation ${a}_{0}$, iterate using the formula:

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

This is a form of Newton Raphson method. Given a function $f \left(x\right)$ for which we want to solve $f \left(x\right) = 0$, start with an approximation ${a}_{0}$ and iterate using:

${a}_{i + 1} = {a}_{i} - f \frac{{a}_{i}}{f ' \left({a}_{i}\right)}$

In our case $f \left(x\right) = {x}^{3} - n$ and $f ' \left(x\right) = 3 {x}^{2}$