Question

question
simplify \\(\sqrt4{1296x^{40}}\\) completely given \\(x > 0\\).

Explanation:

Step1: Simplify the constant term

We know that \(1296 = 6^4\) since \(6\times6\times6\times6 = 1296\). So we can rewrite the fourth - root of the constant part as \(\sqrt[4]{6^4}\). By the property of \(n\) - th roots, \(\sqrt[n]{a^n}=a\) when \(n\) is even and \(a\geq0\). Here \(n = 4\) and \(6\geq0\), so \(\sqrt[4]{6^4}=6\).

Step2: Simplify the variable term

For the variable part \(x^{40}\), we use the property of exponents \((a^m)^n=a^{mn}\) and the property of \(n\) - th roots \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\) (when \(a\geq0\) for even \(n\)). We want to find \(\sqrt[4]{x^{40}}\). Using the formula \(\sqrt[n]{x^m}=x^{\frac{m}{n}}\), with \(n = 4\) and \(m = 40\), we have \(x^{\frac{40}{4}}=x^{10}\). Since \(x>0\), we don't have to worry about the sign of the root (for even roots, if the radicand is non - negative, the root is non - negative).

Step3: Combine the results

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}\) (for \(a\geq0,b\geq0\) and \(n\) a positive integer), we can combine the simplified constant and variable parts. So \(\sqrt[4]{1296x^{40}}=\sqrt[4]{1296}\times\sqrt[4]{x^{40}}\). Substituting the values we found in Step 1 and Step 2, we get \(6\times x^{10}=6x^{10}\).

Answer:

\(6x^{10}\)