How do you find a one-decimal place approximation for $- \sqrt[4]{200}$ $-root4 200$ ?

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How do you find a one-decimal place approximation for $- \sqrt[4]{200}$ $-root4 200$ ?

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Answer 1 · 2015-10-23T22:07:15

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

Explanation:

$3^{4} = 81$ $3^4 = 81$ and $4^{4} = 256$ $4^4 = 256$ , so $\sqrt[4]{200}$ $root(4)(200)$ lies somewhere between $3$ $3$ and $4$ $4$ .

More specifically, $200 \approx 196 = 14^{2}$ $200 ~~ 196 = 14^2$ so $\sqrt[4]{200} \approx \sqrt{14}$ $root(4)(200) ~~ sqrt(14)$ and $\sqrt{14} \approx 4 - \frac{1}{4} = 3.75$ $sqrt(14) ~~ 4-1/4 = 3.75$ , so $\sqrt[4]{200} \approx 3.8$ $root(4)(200) ~~ 3.8$

To find the $4$ $4$ th root of a number $n$ $n$ , we could find its square root and then the square root of that, but instead let's choose a reasonable first approximation $a_{0}$ $a_0$ , then use the following formula to iterate:

$a_{i + 1} = a_{i} + \frac{n - a_{i}^{4}}{4 a_{i}^{3}}$ $a_(i+1) = a_i + (n - a_i^4)/(4a_i^3)$

Let $a_{0} = 3.8$ $a_0 = 3.8$

Then:

$a_{1} = a_{0} + \frac{200 - {3.8}^{4}}{4 \cdot {3.8}^{3}}$ $a_1 = a_0 + (200-3.8^4)/(4*3.8^3)$

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

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

So $3.8$ $3.8$ is a good approximation of $\sqrt[4]{200}$ $root(4)(200)$ to one decimal place and $- 3.8$ $-3.8$ is a one decimal place approximation for $- \sqrt[4]{200}$ $-root(4)(200)$

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How do you find a one-decimal place approximation for $- \sqrt[4]{200}$ $-root4 200$ ?

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