- Identify perfect square factors in the radicands: \(45 = 9 \times 5\) and \(20 = 4 \times 5\).
- Rewrite the square roots as \(\sqrt{45y} = 3\sqrt{5y}\) and \(\sqrt{20y} = 2\sqrt{5y}\).
- Combine the like terms to arrive at the final simplified expression.
- The final result is \(\boxed{5\sqrt{5y}}\).
Explanation
- Problem Analysis
We are tasked with simplifying the expression \(\sqrt{45y} + \sqrt{20y}\) for a positive real number \(y > 0\). To simplify radical expressions of this form, we aim to extract the largest perfect square factor from each radicand. This allows us to combine terms that share the same radical component.
- Factoring the Radicands
First, we factor each radicand to identify perfect square factors.
For \(\sqrt{45y}\), we observe that \(45 = 9 \times 5\). Thus,
\[\sqrt{45y} = \sqrt{9 \times 5y}.\]
For \(\sqrt{20y}\), we observe that \(20 = 4 \times 5\). Thus,
\[\sqrt{20y} = \sqrt{4 \times 5y}.\]
- Simplifying and Combining Like Terms
We apply the property of square roots \(\sqrt{ab} = \sqrt{a}\sqrt{b}\) for \(a, b \ge 0\).
Applying this to our expressions, we get:
\[\sqrt{45y} = \sqrt{9} \times \sqrt{5y} = 3\sqrt{5y}\]
\[\sqrt{20y} = \sqrt{4} \times \sqrt{5y} = 2\sqrt{5y}\]
Now, substitute these back into the original sum:
\[\sqrt{45y} + \sqrt{20y} = 3\sqrt{5y} + 2\sqrt{5y}.\]
- Conclusion
Since both terms now share the same radical part \(\sqrt{5y}\), we can combine them by adding their coefficients:
\[(3 + 2)\sqrt{5y} = 5\sqrt{5y}.\]
Therefore, the simplified form of the expression is \(\boxed{5\sqrt{5y}}\).
Examples
Simplifying radical expressions is a fundamental skill in algebra that appears frequently in geometry and physics. For example, if you are calculating the distance between two points in a coordinate plane where the coordinates involve variables, or when determining the result of physical formulas like the period of a pendulum \(T = 2\pi\sqrt{\frac{L}{g}}\), being able to simplify square root expressions allows for cleaner, more manageable final results. Mastering these manipulations helps in standardizing expressions for further calculations.