- Factor out perfect squares from each radical: \(\sqrt{75z} = 5\sqrt{3z}\) and \(\sqrt{12z} = 2\sqrt{3z}\).
- Substitute these back into the original expression: \(5\sqrt{3z} - 2\sqrt{3z}\).
- Subtract the coefficients to obtain the final simplified result: \(\boxed{3\sqrt{3z}}\).
Explanation
- Analysis and Objective
We are given the radical expression \(\sqrt{75z} - \sqrt{12z}\) where \(z\) is a positive real number. Our objective is to simplify this expression by combining like radical terms. The key property we will use is that for any non-negative numbers \(a\) and \(b\), \(\sqrt{ab} = \sqrt{a}\sqrt{b}\). Additionally, we will factor out the largest perfect square from each radicand to simplify the square roots.
- Simplifying Radical Terms
First, we simplify each radical term individually:
\begin{itemize}
\item For the first term, \(\sqrt{75z}\), we can factor \(75\) into \(25 \times 3\):
\[\sqrt{75z} = \sqrt{25 \times 3 \times z} = \sqrt{25} \times \sqrt{3z} = 5\sqrt{3z}\]
\item For the second term, \(\sqrt{12z}\), we can factor \(12\) into \(4 \times 3\):
\[\sqrt{12z} = \sqrt{4 \times 3 \times z} = \sqrt{4} \times \sqrt{3z} = 2\sqrt{3z}\]
\end{itemize}
- Combining Terms and Final Simplification
Now, substitute these simplified forms back into the original expression:
\[\sqrt{75z} - \sqrt{12z} = 5\sqrt{3z} - 2\sqrt{3z}\]
Since both terms contain the common factor \(\sqrt{3z}\), we can use the distributive property to subtract the coefficients:
\[5\sqrt{3z} - 2\sqrt{3z} = (5 - 2)\sqrt{3z} = 3\sqrt{3z}\]
Thus, the simplified form of the expression is \(3\sqrt{3z}\).
Examples
Radical simplification is useful in various engineering and physics fields when calculating magnitudes or working with formulas involving square roots. For instance, if you are designing a structural component where the dimensions are proportional to \(\sqrt{75z}\) and \(\sqrt{12z}\), knowing the difference allows you to simplify your formulas for easier measurement and manufacturing. Similarly, in electrical circuits, simplifying expressions with radicals can help in calculating total impedance more efficiently.