Question

simplify the following: \\(\frac{\sqrt4{200}}{\sqrt{128}}\\)

Explanation:

Step1: Simplify the numerator (4th root of 200)

First, factorize 200. \(200 = 2^3\times5^2\). The 4th root of 200 is \(\sqrt[4]{2^3\times5^2}\). We can rewrite this as \(2^{\frac{3}{4}}\times5^{\frac{2}{4}} = 2^{\frac{3}{4}}\times5^{\frac{1}{2}}\).

Step2: Simplify the denominator (square root of 128)

Factorize 128. \(128 = 2^7\). The square root of 128 is \(\sqrt{2^7}=2^{\frac{7}{2}}\).

Step3: Divide the numerator by the denominator

Now we have \(\frac{2^{\frac{3}{4}}\times5^{\frac{1}{2}}}{2^{\frac{7}{2}}}\). Using the rule of exponents \(\frac{a^m}{a^n}=a^{m - n}\) for the base 2:
\(2^{\frac{3}{4}-\frac{7}{2}}\times5^{\frac{1}{2}}\)
Calculate the exponent of 2: \(\frac{3}{4}-\frac{7}{2}=\frac{3 - 14}{4}=-\frac{11}{4}\)
So we have \(2^{-\frac{11}{4}}\times5^{\frac{1}{2}}\). We can rewrite \(2^{-\frac{11}{4}}\) as \(\frac{1}{2^{\frac{11}{4}}}=\frac{1}{\sqrt[4]{2^{11}}}=\frac{1}{\sqrt[4]{2^8\times2^3}}=\frac{1}{2^2\times\sqrt[4]{8}}=\frac{1}{4\sqrt[4]{8}}\)
And \(5^{\frac{1}{2}}=\sqrt{5}\)
So the expression becomes \(\frac{\sqrt{5}}{4\sqrt[4]{8}}\). But we can also rationalize or rewrite in a combined root form. Alternatively, let's express all exponents with a common denominator.
Another way: Let's rewrite the original fraction as \(\frac{\sqrt[4]{200}}{\sqrt{128}}=\sqrt[4]{\frac{200}{128^2}}\) (since \(\frac{\sqrt[4]{a}}{\sqrt{b}}=\sqrt[4]{\frac{a}{b^2}}\) because \(\sqrt{b}=b^{\frac{1}{2}}\) and \(\frac{a^{\frac{1}{4}}}{b^{\frac{1}{2}}}=a^{\frac{1}{4}}b^{-\frac{1}{2}}=(a\times b^{-2})^{\frac{1}{4}}=\sqrt[4]{\frac{a}{b^2}}\))
Calculate \(128^2=(2^7)^2 = 2^{14}\)
So \(\frac{200}{2^{14}}=\frac{2^3\times5^2}{2^{14}}=2^{-11}\times5^2\)
Then \(\sqrt[4]{2^{-11}\times5^2}=2^{-\frac{11}{4}}\times5^{\frac{2}{4}}=2^{-\frac{11}{4}}\times5^{\frac{1}{2}}\) as before.
Alternatively, let's simplify the fraction \(\frac{\sqrt[4]{200}}{\sqrt{128}}\) by first simplifying the fraction inside the roots (after adjusting the roots to the same index). The least common multiple of 4 and 2 is 4. So rewrite \(\sqrt{128}\) as \(\sqrt[4]{128^2}=\sqrt[4]{16384}\)
And \(\sqrt[4]{200}\) remains as is. So now we have \(\sqrt[4]{\frac{200}{16384}}=\sqrt[4]{\frac{25}{2048}}\) (dividing numerator and denominator by 8)
\(2048 = 2^{11}\), \(25 = 5^2\)
So \(\sqrt[4]{\frac{5^2}{2^{11}}}=\frac{5^{\frac{2}{4}}}{2^{\frac{11}{4}}}=\frac{5^{\frac{1}{2}}}{2^{\frac{11}{4}}}=\frac{\sqrt{5}}{2^2\times\sqrt[4]{2^3}}=\frac{\sqrt{5}}{4\sqrt[4]{8}}\)
We can also rationalize the denominator by multiplying numerator and denominator by \(\sqrt[4]{2}\):
\(\frac{\sqrt{5}\times\sqrt[4]{2}}{4\sqrt[4]{8}\times\sqrt[4]{2}}=\frac{\sqrt{5}\times\sqrt[4]{2}}{4\sqrt[4]{16}}=\frac{\sqrt{5}\times\sqrt[4]{2}}{4\times2}=\frac{\sqrt{5}\times\sqrt[4]{2}}{8}\)
Or rewrite \(\sqrt[4]{2}\) as \(2^{\frac{1}{4}}\) and \(\sqrt{5}\) as \(5^{\frac{1}{2}}\), so the expression is \(\frac{5^{\frac{1}{2}}\times2^{\frac{1}{4}}}{8}\)

But maybe a simpler way: Let's start over with prime factorization for both numerator and denominator roots.

Numerator: \(\sqrt[4]{200}=\sqrt[4]{2^3\times5^2}\)

Denominator: \(\sqrt{128}=\sqrt{2^7}=2^{\frac{7}{2}}\)

So the fraction is \(\frac{2^{\frac{3}{4}}\times5^{\frac{2}{4}}}{2^{\frac{7}{2}}}=2^{\frac{3}{4}-\frac{14}{4}}\times5^{\frac{1}{2}}=2^{-\frac{11}{4}}\times\sqrt{5}=\frac{\sqrt{5}}{2^{\frac{11}{4}}}\)

\(2^{\frac{11}{4}}=2^{2 + \frac{3}{4}}=4\times2^{\frac{3}{4}}=4\sqrt[4]{8}\)

So \(\frac{\sqrt{5}}{4\sqrt[4]{8}}\) can be written as \(\frac{\sqrt{5}\times\sqrt[4]{2}}{4\sqrt[4]{8}\times\sqrt[4]{2}}=\frac{\sqrt{5}\times\sqrt[4]{2}}{4\times2}=\frac{\sqrt{5}\times\sqrt[4]{2}}{8}\)

Alternatively, if we want to express in terms of square roots and fourth roots combined, but the simplest form (rationalized denominator) is \(\frac{\sqrt{5}\times\sqrt[4]{2}}{8}\) or \(\frac{\sqrt[4]{50}}{8}\) (since \(\sqrt{5}\times\sqrt[4]{2}=\sqrt[4]{25}\times\sqrt[4]{2}=\sqrt[4]{50}\))

Let's check: \(\sqrt[4]{50}\approx2.659\), \(\sqrt{5}\approx2.236\), \(\sqrt[4]{2}\approx1.189\), \(2.236\times1.189\approx2.659\), and \(8\times2.659\approx21.27\), and the original expression \(\frac{\sqrt[4]{200}}{\sqrt{128}}\approx\frac{3.737}{11.313}\approx0.330\), and \(\frac{\sqrt[4]{50}}{8}\approx\frac{2.659}{8}\approx0.332\) (close due to rounding). So that works.

So the simplified form is \(\frac{\sqrt[4]{50}}{8}\) or \(\frac{\sqrt{5}\sqrt[4]{2}}{8}\)

Answer:

\(\frac{\sqrt[4]{50}}{8}\) (or equivalent forms like \(\frac{\sqrt{5}\sqrt[4]{2}}{8}\))