Question

write the following expression in simplified radical form.
\sqrt4{81x^{10}z^{17}}
assume that all of the variables in the expression represent positive real numbers.

Explanation:

Step1: Simplify the constant term

We know that \(81 = 3^4\), so \(\sqrt[4]{81}=\sqrt[4]{3^4} = 3\).

Step2: Simplify the \(x\) - term

For the variable \(x\) with exponent \(10\), we use the property of radicals \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\) and also express \(m\) as a multiple of \(n\) plus a remainder. We can write \(10 = 4\times2+2\). So \(\sqrt[4]{x^{10}}=\sqrt[4]{x^{4\times2 + 2}}=\sqrt[4]{(x^{4})^{2}\times x^{2}}=x^{2}\sqrt[4]{x^{2}}\) (since \(\sqrt[4]{(x^{4})^{2}}=x^{2}\) as \(x>0\)).

Step3: Simplify the \(z\) - term

For the variable \(z\) with exponent \(17\), we write \(17=4\times4 + 1\). So \(\sqrt[4]{z^{17}}=\sqrt[4]{z^{4\times4+1}}=\sqrt[4]{(z^{4})^{4}\times z^{1}}=z^{4}\sqrt[4]{z}\) (since \(\sqrt[4]{(z^{4})^{4}} = z^{4}\) as \(z>0\)).

Step4: Combine all the simplified terms

Now, we multiply the simplified constant, \(x\) - term and \(z\) - term together:
\(\sqrt[4]{81x^{10}z^{17}}=\sqrt[4]{81}\times\sqrt[4]{x^{10}}\times\sqrt[4]{z^{17}}=3\times x^{2}\sqrt[4]{x^{2}}\times z^{4}\sqrt[4]{z}\)
\(= 3x^{2}z^{4}\sqrt[4]{x^{2}z}\)
We can also simplify \(\sqrt[4]{x^{2}}\) as \(x^{\frac{2}{4}}=x^{\frac{1}{2}}=\sqrt{x}\), but in radical form for the fourth root, we can rewrite \(x^{2}=x^{4\times\frac{1}{2}}\), but actually, since we want to keep it in fourth - root form, we note that \(x^{2}\) under the fourth root can be written as \(x^{2}=x^{4\times0.5}\), but a better way is to use the property \(\sqrt[4]{x^{2}}=\sqrt{\sqrt{x^{2}}}=\sqrt{x}\) (since \(x>0\)), but we can also keep it as \(\sqrt[4]{x^{2}}\). However, we can rewrite \(x^{10}=x^{8 + 2}=(x^{2})^{4}\times x^{2}\) and \(z^{17}=(z^{4})^{4}\times z\) more accurately. Let's re - do the exponent handling:

For \(x^{10}\), \(\frac{10}{4}=\frac{5}{2}=2+\frac{1}{2}\), so \(\sqrt[4]{x^{10}}=x^{\frac{10}{4}}=x^{2+\frac{1}{2}}=x^{2}\sqrt[4]{x^{2}}\) (because \(x^{\frac{10}{4}}=x^{2+\frac{2}{4}}=x^{2}\times x^{\frac{2}{4}}=x^{2}\sqrt[4]{x^{2}}\)).

For \(z^{17}\), \(\frac{17}{4}=4+\frac{1}{4}\), so \(\sqrt[4]{z^{17}}=z^{\frac{17}{4}}=z^{4+\frac{1}{4}}=z^{4}\sqrt[4]{z}\).

So \(\sqrt[4]{81x^{10}z^{17}}=\sqrt[4]{3^{4}}\times\sqrt[4]{x^{10}}\times\sqrt[4]{z^{17}}=3\times x^{2}\sqrt[4]{x^{2}}\times z^{4}\sqrt[4]{z}\)

We can also write \(\sqrt[4]{x^{2}}=\sqrt{x}\) (since \(\sqrt[4]{x^{2}}=(x^{2})^{\frac{1}{4}}=x^{\frac{1}{2}}=\sqrt{x}\)), but if we want to keep it in fourth - root form, we can note that \(x^{2}=x^{4\times\frac{1}{2}}\), but the standard simplified radical form for \(\sqrt[4]{x^{2}}\) can be written as \(\sqrt{x}\) (since for non - negative \(x\), \(\sqrt[4]{x^{2}}=\sqrt{x}\)). However, let's use the exponent division properly:

\(\sqrt[4]{81x^{10}z^{17}}=3\times x^{\frac{10}{4}}\times z^{\frac{17}{4}}=3\times x^{2+\frac{2}{4}}\times z^{4+\frac{1}{4}}=3x^{2}z^{4}\times x^{\frac{1}{2}}\times z^{\frac{1}{4}}=3x^{2}z^{4}\sqrt[4]{x^{2}z}\) (since \(x^{\frac{1}{2}}=\sqrt{x}\) and \(z^{\frac{1}{4}}=\sqrt[4]{z}\), and \(x^{2}z^{4}\) is the part that comes out of the radical).

Another way:

We know that \(\sqrt[4]{ab}=\sqrt[4]{a}\times\sqrt[4]{b}\) for \(a,b\geq0\).

\(\sqrt[4]{81x^{10}z^{17}}=\sqrt[4]{81}\times\sqrt[4]{x^{10}}\times\sqrt[4]{z^{17}}\)

\(=3\times\sqrt[4]{x^{8 + 2}}\times\sqrt[4]{z^{16+1}}\)

\(=3\times\sqrt[4]{x^{8}\times x^{2}}\times\sqrt[4]{z^{16}\times z}\)

\(=3\times x^{2}\times\sqrt[4]{x^{2}}\times z^{4}\times\sqrt[4]{z}\)

\(=3x^{2}z^{4}\sqrt[4]{x^{2}z}\)

We can also simplify \(\sqrt[4]{x^{2}}\) as \(x^{\frac{2}{4}}=x^{\frac{1}{2}}=\sqrt{x}\), but the simplified radical form can be written as \(3x^{2}z^{4}\sqrt[4]{x^{2}z}\) or if we want to write \(\sqrt[4]{x^{2}}\) as \(\sqrt{x}\), we can rewrite it as \(3x^{2}z^{4}\sqrt{x}\sqrt[4]{z}\), but the standard form is to have the radical with the smallest possible index. Since \(x^{2}\) is a perfect square, \(\sqrt[4]{x^{2}}=\sqrt{x}\), so:

\(3x^{2}z^{4}\sqrt{x}\sqrt[4]{z}=3x^{2}z^{4}\sqrt[4]{x^{2}z}\) (because \(\sqrt{x}=\sqrt[4]{x^{2}}\))

Answer:

\(3x^{2}z^{4}\sqrt[4]{x^{2}z}\)