- Simplify the first term \(\sqrt{3 v u^2}\) to \(u \sqrt{3v}\), giving \(9 u v^2 \sqrt{3v}\).
- Simplify the second term \(\sqrt{48 v^5}\) to \(4 v^2 \sqrt{3v}\), giving \(4 u v^2 \sqrt{3v}\).
- Subtract the second term from the first to get \((9-4) u v^2 \sqrt{3v}\).
- The final simplified expression is \(\boxed{5 u v^2 \sqrt{3v}}\).
Explanation
- Problem Analysis
To simplify the given expression \(9 v^2 \sqrt{3 v u^2} - u \sqrt{48 v^5}\), where \(u > 0\) and \(v > 0\), we need to simplify each radical term individually and then combine like terms. This process involves identifying and factoring out perfect square expressions from inside the square root. The variables are given as positive real numbers, which allows us to use the property \(\sqrt{x^2} = |x| = x\) for any \(x > 0\).
- Simplifying the First Term
We start by simplifying the first term: \(9 v^2 \sqrt{3 v u^2}\).
Since \(u > 0\), we have \(\sqrt{u^2} = u\). Thus, we can pull \(u\) out of the square root:
\[9 v^2 \sqrt{3 v u^2} = 9 v^2 \cdot u \cdot \sqrt{3v} = 9 u v^2 \sqrt{3v}.\]
- Simplifying the Second Term
Next, we simplify the second term: \(u \sqrt{48 v^5}\).
We factor the radicand \(48 v^5\) into perfect squares. We know \(48 = 16 \times 3 = 4^2 \times 3\) and \(v^5 = v^4 \times v = (v^2)^2 \times v\). Therefore:
\[\sqrt{48 v^5} = \sqrt{16 \times 3 \times v^4 \times v} = \sqrt{16} \cdot \sqrt{v^4} \cdot \sqrt{3v} = 4 \cdot v^2 \cdot \sqrt{3v}.\]
Multiplying this by the leading factor \(u\), we get:
\[u \sqrt{48 v^5} = u \cdot (4 v^2 \sqrt{3v}) = 4 u v^2 \sqrt{3v}.\]
- Combining Terms and Final Answer
Finally, we combine the two simplified terms:
\[9 u v^2 \sqrt{3v} - 4 u v^2 \sqrt{3v}.\]
Since both terms share the common factor \(u v^2 \sqrt{3v}\), we can subtract the coefficients:
\[(9 - 4) u v^2 \sqrt{3v} = 5 u v^2 \sqrt{3v}.\]
The simplified expression is \(\boxed{5 u v^2 \sqrt{3v}}\).
Examples
Simplifying radical expressions is essential in fields like physics and engineering, where you often encounter formulas involving square roots to calculate distances, velocities, or resistances. For example, if the velocity of an object is described by a function \(V(u, v) = 9 v^2 \sqrt{3 v u^2} - u \sqrt{48 v^5}\) (measured in meters per second), simplifying this expression allows you to quickly determine how the velocity scales with changes in parameters \(u\) and \(v\), making it much easier to model real-world movement.