Question

add or subtract. assume that all variables represent positive real numbers.
$-2sqrt4{x^{6}} + 3sqrt4{16x^{6}}$

$-2sqrt4{x^{6}} + 3sqrt4{16x^{6}} = \square$
(type an exact answer, using radicals as needed. simplify your answer.)

Explanation:

Step1: Simplify the fourth - root terms

First, simplify \(\sqrt[4]{16x^{6}}\). We know that \(16 = 2^{4}\) and \(x^{6}=x^{4}\cdot x^{2}\). So, \(\sqrt[4]{16x^{6}}=\sqrt[4]{2^{4}\cdot x^{4}\cdot x^{2}}\).
Using the property of radicals \(\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}\) (\(a\geq0,b\geq0\)) and \(\sqrt[n]{a^{n}} = a\) (for \(a\geq0\) and \(n\) even), we have \(\sqrt[4]{2^{4}\cdot x^{4}\cdot x^{2}}=\sqrt[4]{2^{4}}\cdot\sqrt[4]{x^{4}}\cdot\sqrt[4]{x^{2}} = 2x\sqrt[4]{x^{2}}\).
Also, simplify \(\sqrt[4]{x^{6}}\). Since \(x^{6}=x^{4}\cdot x^{2}\), then \(\sqrt[4]{x^{6}}=\sqrt[4]{x^{4}\cdot x^{2}}=\sqrt[4]{x^{4}}\cdot\sqrt[4]{x^{2}}=x\sqrt[4]{x^{2}}\).

Step2: Substitute the simplified terms back into the original expression

The original expression is \(- 2\sqrt[4]{x^{6}}+3\sqrt[4]{16x^{6}}\). Substituting the simplified forms:
\(-2\sqrt[4]{x^{6}}+3\sqrt[4]{16x^{6}}=-2x\sqrt[4]{x^{2}}+3\times(2x\sqrt[4]{x^{2}})\)
\(=-2x\sqrt[4]{x^{2}} + 6x\sqrt[4]{x^{2}}\)

Step3: Combine like terms

Combine the terms with \(x\sqrt[4]{x^{2}}\). Using the rule \(ax + bx=(a + b)x\) (here \(x\) is \(x\sqrt[4]{x^{2}}\)), we have \((-2 + 6)x\sqrt[4]{x^{2}}=4x\sqrt[4]{x^{2}}\).
We can also rewrite \(\sqrt[4]{x^{2}}\) as \(x^{\frac{2}{4}}=x^{\frac{1}{2}}=\sqrt{x}\), but we can leave it in the fourth - root form. Another way to simplify \(\sqrt[4]{x^{2}}\) is to note that \(\sqrt[4]{x^{2}}=\sqrt{\sqrt{x^{2}}}=\sqrt{x}\) (since \(x>0\)), but we can also keep it as \(\sqrt[4]{x^{2}}\). However, let's go back to the radical simplification.
Wait, we made a mistake in the first step. Let's re - simplify \(\sqrt[4]{x^{6}}\) and \(\sqrt[4]{16x^{6}}\) correctly.
Correct simplification:
For \(\sqrt[4]{x^{6}}\), \(x^{6}=x^{4 + 2}=x^{4}\cdot x^{2}\), so \(\sqrt[4]{x^{6}}=\sqrt[4]{x^{4}\cdot x^{2}}=x\sqrt[4]{x^{2}}\), and \(\sqrt[4]{x^{2}}=\sqrt{x}\) (since \(x>0\), \(\sqrt[4]{x^{2}}=(x^{2})^{\frac{1}{4}}=x^{\frac{1}{2}}=\sqrt{x}\)). So \(\sqrt[4]{x^{6}}=x\sqrt{x}\).
For \(\sqrt[4]{16x^{6}}\), \(16 = 2^{4}\), \(x^{6}=x^{4}\cdot x^{2}\), so \(\sqrt[4]{16x^{6}}=\sqrt[4]{2^{4}\cdot x^{4}\cdot x^{2}}=2x\sqrt[4]{x^{2}}=2x\sqrt{x}\).
Now, substitute back into the original expression:
\(-2\sqrt[4]{x^{6}}+3\sqrt[4]{16x^{6}}=-2x\sqrt{x}+3\times(2x\sqrt{x})\)
\(=-2x\sqrt{x}+6x\sqrt{x}\)

Step4: Combine like terms

Using the rule \(ax+bx=(a + b)x\) where \(a=-2\), \(b = 6\) and \(x=x\sqrt{x}\), we get \((-2 + 6)x\sqrt{x}=4x\sqrt{x}\).
Or, if we keep it in the fourth - root form:
We know that \(x^{2}=x^{2}\), so \(\sqrt[4]{x^{2}}=x^{\frac{2}{4}}=x^{\frac{1}{2}}\), and the original expression after substitution:
\(-2x\sqrt[4]{x^{2}}+6x\sqrt[4]{x^{2}}=( - 2 + 6)x\sqrt[4]{x^{2}}=4x\sqrt[4]{x^{2}}\). But \(\sqrt[4]{x^{2}}=\sqrt{x}\) (since \(x>0\)), so \(4x\sqrt[4]{x^{2}} = 4x\sqrt{x}\).
Another way:
Let's start over.
\(\sqrt[4]{x^{6}}=x^{\frac{6}{4}}=x^{\frac{3}{2}}=x\sqrt{x}\) (since \(x^{\frac{3}{2}}=x^{1+\frac{1}{2}}=x\sqrt{x}\))
\(\sqrt[4]{16x^{6}}=\sqrt[4]{16}\cdot\sqrt[4]{x^{6}}=2\cdot x^{\frac{6}{4}}=2x^{\frac{3}{2}}=2x\sqrt{x}\)
So the original expression \(-2\sqrt[4]{x^{6}}+3\sqrt[4]{16x^{6}}=-2x\sqrt{x}+3\times(2x\sqrt{x})=-2x\sqrt{x}+6x\sqrt{x}\)
Combine like terms: \((-2 + 6)x\sqrt{x}=4x\sqrt{x}\)

Answer:

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