Question

question
simplify \\(\sqrt4{1296x^{36}}\\) completely given \\(x > 0\\).

answer attempt 1 out of 4

Explanation:

Step1: Simplify the constant term

We know that \(1296 = 6^4\), so \(\sqrt[4]{1296}=\sqrt[4]{6^4} = 6\) (since for any non - negative real number \(a\) and positive integer \(n\), \(\sqrt[n]{a^n}=a\) when \(n\) is even and \(a\geq0\), and here \(a = 6\), \(n=4\) and \(6>0\)).

Step2: Simplify the variable term

For the term \(x^{36}\), using the property of radicals \(\sqrt[n]{a^m}=a^{\frac{m}{n}}\) (where \(a\geq0\), \(m\) and \(n\) are positive integers), we have \(\sqrt[4]{x^{36}}=x^{\frac{36}{4}}=x^{9}\) (since \(x>0\), we don't have to consider the absolute value here).

Step3: Combine the results

Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (where \(a\geq0\), \(b\geq0\) and \(n\) is a positive integer), we get \(\sqrt[4]{1296x^{36}}=\sqrt[4]{1296}\cdot\sqrt[4]{x^{36}}=6\cdot x^{9}=6x^{9}\).

Answer:

\(6x^{9}\)