Question

simplify the fourth - root expression (sqrt4{64x^{4}})

Explanation:

Step1: Simplify the constant term

We know that \(64 = 2^6\), but we need to express it as a fourth - power or a product of a fourth - power and another number. We can rewrite \(64\) as \(16\times4=2^{4}\times4\). So, \(\sqrt[4]{64x^{4}}=\sqrt[4]{2^{4}\times4\times x^{4}}\)

Step2: Use the property of fourth - roots \(\sqrt[4]{ab}=\sqrt[4]{a}\cdot\sqrt[4]{b}\) (\(a\geq0,b\geq0\))

According to the property \(\sqrt[4]{ab}=\sqrt[4]{a}\cdot\sqrt[4]{b}\), we have \(\sqrt[4]{2^{4}\times4\times x^{4}}=\sqrt[4]{2^{4}}\cdot\sqrt[4]{x^{4}}\cdot\sqrt[4]{4}\)

Step3: Simplify \(\sqrt[4]{2^{4}}\) and \(\sqrt[4]{x^{4}}\)

We know that for any real number \(a\), \(\sqrt[4]{a^{4}}=\vert a\vert\) when \(n = 4\) (even root). But if we assume \(x\geq0\) (since if we are dealing with a simplified form in the context of real - valued functions and the expression is likely defined for non - negative \(x\) in a basic algebra context), \(\sqrt[4]{x^{4}}=x\) and \(\sqrt[4]{2^{4}} = 2\)

Step4: Simplify \(\sqrt[4]{4}\)

We can rewrite \(4\) as \(2^{2}\), so \(\sqrt[4]{4}=\sqrt[4]{2^{2}}=2^{\frac{2}{4}}=2^{\frac{1}{2}}=\sqrt{2}\)

Step5: Combine the terms

Multiply the simplified terms together: \(2\times x\times\sqrt{2}=2\sqrt{2}x\) (if we consider the principal root and \(x\geq0\)). Another way: We can also rewrite \(64 = 2^{6}=2^{4 + 2}=2^{4}\times2^{2}\), so \(\sqrt[4]{64x^{4}}=\sqrt[4]{2^{4}\times2^{2}\times x^{4}}=\sqrt[4]{2^{4}}\times\sqrt[4]{x^{4}}\times\sqrt[4]{2^{2}}=2\times x\times2^{\frac{2}{4}}=2x\times\sqrt{2}\)

If we want to express it in a more "simplified" radical form without a square root inside the fourth - root, we can also note that \(64=4\times16 = 4\times2^{4}\), then \(\sqrt[4]{64x^{4}}=\sqrt[4]{4\times2^{4}\times x^{4}}=\sqrt[4]{2^{4}}\times\sqrt[4]{x^{4}}\times\sqrt[4]{4}=2x\sqrt[4]{4}=2x\sqrt{2}\) (since \(\sqrt[4]{4}=\sqrt{2}\))

Answer:

\(2\sqrt{2}x\) (or \(2x\sqrt[4]{4}\), but \(2\sqrt{2}x\) is a more simplified form)