# What is the square root of 3.4?

Sep 18, 2015

$\sqrt{3.4} = \frac{\sqrt{85}}{5} \approx 1.8439$

#### Explanation:

If $a , b \ge 0$ then $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$

If $a \ge 0$ then $\sqrt{{a}^{2}} = a$

So:

$\sqrt{3.4} = \sqrt{\frac{85}{25}} = \frac{\sqrt{85}}{\sqrt{25}} = \frac{\sqrt{85}}{5}$

Here I have chosen to express $3.4$ as $\frac{85}{25}$ to get the smallest whole value for the denominator.

I could also have written $\sqrt{3.4} = \sqrt{\frac{17}{5}} = \frac{\sqrt{17}}{\sqrt{5}}$

If I wanted to calculate an approximation for $\sqrt{3.4}$ by hand then I would probably have used this instead:

$\sqrt{3.4} = \sqrt{\frac{340}{100}} = \frac{\sqrt{340}}{10}$

Then I would work out an approximation for $\sqrt{340}$ and divide by $10$.

For example: ${18}^{2} = 324$ and ${19}^{2} = 361$, so:

$18 = \sqrt{324} < \sqrt{340} < \sqrt{361} = 19$

Using a Newton Raphson type method to find $\sqrt{340}$, I might chose $18.5 = \frac{37}{2}$ as my first approximation ${a}_{0}$, then make better approximations using the formula:

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

where $n = 340$.

Then

${a}_{1} = \frac{{37}^{2} + 340 \cdot {2}^{2}}{2 \cdot 37 \cdot 2} = \frac{1369 + 1360}{148}$

$= \frac{2729}{148} \approx 18.439$

So $\sqrt{340} \approx 18.439$ and $\sqrt{3.4} \approx 1.8439$