# How do you find a one-decimal place approximation for -root4 200?

Oct 23, 2015

Use some reasonings about ${3}^{4}$, ${4}^{4}$ and $\sqrt{200}$ to choose approximation $3.8$, then use one step of Newton's method to refine the approximation, finding that $3.8$ is good to one decimal place already.

#### Explanation:

${3}^{4} = 81$ and ${4}^{4} = 256$, so $\sqrt{200}$ lies somewhere between $3$ and $4$.

More specifically, $200 \approx 196 = {14}^{2}$ so $\sqrt{200} \approx \sqrt{14}$ and $\sqrt{14} \approx 4 - \frac{1}{4} = 3.75$, so $\sqrt{200} \approx 3.8$

To find the $4$th root of a number $n$, we could find its square root and then the square root of that, but instead let's choose a reasonable first approximation ${a}_{0}$, then use the following formula to iterate:

${a}_{i + 1} = {a}_{i} + \frac{n - {a}_{i}^{4}}{4 {a}_{i}^{3}}$

Let ${a}_{0} = 3.8$

Then:

${a}_{1} = {a}_{0} + \frac{200 - {3.8}^{4}}{4 \cdot {3.8}^{3}}$

$= 3.8 + \frac{200 - 208.5136}{219.488}$

$= 3.8 - \frac{8.5136}{219.488} \approx 3.76$

So $3.8$ is a good approximation of $\sqrt{200}$ to one decimal place and $- 3.8$ is a one decimal place approximation for $- \sqrt{200}$