# What is the square root of 67.98?

Oct 13, 2015

$67.98 = \frac{2 \cdot 3 \cdot 11 \cdot 103}{100}$, so the simplest algebraic form is:

$\sqrt{67.98} = \frac{\sqrt{6798}}{10} \approx 8.245$

#### Explanation:

To calculate an approximation for $\sqrt{67.98}$, find an approximation for $\sqrt{6798}$ and divide by $10$

${80}^{2} = 6400 < 6798 < 8100 = {90}^{2}$

To find the square root of a number $n$, you can choose a first approximation ${a}_{0}$ and iterate using the formula:

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

I prefer to work with integers, so I express ${a}_{0} = {p}_{0} / {q}_{0}$ and iterate using the formulae:

${p}_{i + 1} = {p}_{i}^{2} + n {q}_{i}^{2}$

${q}_{i + 1} = 2 {p}_{i} {q}_{i}$

Then ${p}_{i} / {q}_{i}$ is the same as ${a}_{i}$.

If the resulting ${p}_{i + 1}$ and ${q}_{i + 1}$ have a common factor larger than $1$ then divide both by that before the next iteration.

Let $n = 6798$, ${p}_{0} = 80$, ${q}_{0} = 1$

Then:

${p}_{1} = {p}_{0}^{2} + n {q}_{0}^{2} = {80}^{2} + 6798 \cdot {1}^{2} = 6400 + 6798 = 13198$

${q}_{1} = 2 {p}_{0} {q}_{0} = 2 \cdot 80 \cdot 1 = 160$

These are both divisible by $2$, so do that to get:

${p}_{1 a} = {p}_{1} / 2 = 6599$

${q}_{1 a} = {q}_{1} / 2 = 80$

Next iteration:

${p}_{2} = {p}_{1 a}^{2} + n {q}_{1 a}^{2} = {6599}^{2} + 6798 \cdot {80}^{2} = 43546801 + 43507200 = 87054001$

${q}_{2} = 2 {p}_{1 a} {q}_{1 a} = 2 \cdot 6599 \cdot 80 = 1055840$

Stopping here, we get:

$\sqrt{6798} \approx \frac{87054001}{1055840} \approx 82.44999 \approx 82.45$

So

$\sqrt{67.98} \approx 8.245$

For an alternative method of finding that $\sqrt{6798} \approx \frac{1645}{20} = 82.45$ see:
http://socratic.org/questions/given-an-integer-n-is-there-an-efficient-way-to-find-integers-p-q-such-that-abs-#176764