# How do you find a one-decimal place approximation for sqrt 30?

Oct 18, 2015

$30$ is about half way between $25 = {5}^{2}$ and $36 = {6}^{2}$, so $5.5$ might be a good approximation.

Check ${5.5}^{2} = 30.25$ is pretty close, so yes.

#### Explanation:

Halfway between $25 = {5}^{2}$ and $36 = {6}^{2}$ is $30.5$ not $30$, but the parabola is also curved below the line joining $\left(5 , 25\right)$ and $\left(6 , 36\right)$, so $5.5$ is a better approximation for $\sqrt{30}$ than for $\sqrt{31}$.

If you want more accuracy use something like a Newton Raphson type method:

To find approximations for $\sqrt{n}$, pick a starting approximation ${a}_{0}$ and iterate using the formula:

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

So in our case we might start with ${a}_{0} = 5.5$, then improve it by applying the formula:

${a}_{1} = \frac{{a}_{0}^{2} + n}{2 {a}_{0}} = \frac{{5.5}^{2} + 30}{2 \cdot 5.5} = \frac{30.25 + 30}{11} = \frac{60.25}{11} \approx 5.477$