Question

simplify.
\sqrt4{625w^{28}}
assume that the variable represents a positive real number.

Explanation:

Step1: Recall the property of nth roots

For a non - negative real number \(a\) and positive integer \(n\), \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (product rule of radicals), and \(\sqrt[n]{a^{m}} = a^{\frac{m}{n}}\) when \(a\geq0\). Also, we know that \(625 = 5^{4}\) and we have the fourth root of \(625w^{28}\), so we can apply the product rule of radicals: \(\sqrt[4]{625w^{28}}=\sqrt[4]{625}\cdot\sqrt[4]{w^{28}}\)

Step2: Simplify \(\sqrt[4]{625}\)

Since \(625 = 5\times5\times5\times5=5^{4}\), then \(\sqrt[4]{625}=\sqrt[4]{5^{4}}\). Using the property \(\sqrt[n]{a^{n}}=a\) (for \(a\geq0\) and \(n\) is a positive integer), we get \(\sqrt[4]{5^{4}} = 5\)

Step3: Simplify \(\sqrt[4]{w^{28}}\)

Using the property \(\sqrt[n]{a^{m}}=a^{\frac{m}{n}}\), here \(n = 4\) and \(m = 28\). So \(\sqrt[4]{w^{28}}=w^{\frac{28}{4}}\). Calculating the exponent \(\frac{28}{4}=7\), so \(\sqrt[4]{w^{28}}=w^{7}\)

Step4: Multiply the results

Now, multiply the results of the two fourth - roots: \(\sqrt[4]{625}\cdot\sqrt[4]{w^{28}}=5\times w^{7}=5w^{7}\)

Answer:

\(5w^{7}\)