- Distribute \(3 \sqrt{10}\) into the parentheses to get \(3 \sqrt{10} \cdot y^2 \sqrt{4} + 3 \sqrt{10} \cdot \sqrt{8y}\).
- Simplify the first part to \(6 y^2 \sqrt{10}\) because \(\sqrt{4} = 2\).
- Simplify the second part to \(12 \sqrt{5y}\) by calculating \(3 \sqrt{80y} = 3 \cdot 4 \sqrt{5y}\).
- The final simplified expression is \(6 y^2 \sqrt{10} + 12 \sqrt{5y}\).
Explanation
- Simplifying the Expression
To simplify the expression \(3 \sqrt{10} \left(y^2 \sqrt{4} + \sqrt{8 y}\right)\) given the condition \(y \geq 0\), we begin by applying the distributive property to multiply the term outside the parentheses by each term inside:
\[3 \sqrt{10} \cdot y^2 \sqrt{4} + 3 \sqrt{10} \cdot \sqrt{8 y}.\]
Next, we simplify each term individually:
- For the first term, we know \(\sqrt{4} = 2\), so:
\[3 \sqrt{10} \cdot y^2 \cdot 2 = 6 y^2 \sqrt{10}.\]
- For the second term, we use the properties of radicals \(\sqrt{a} \sqrt{b} = \sqrt{ab}\):
\[3 \sqrt{10} \cdot \sqrt{8 y} = 3 \sqrt{80 y}.\]
Factoring the radical \(\sqrt{80 y}\) to simplify it, we find:
\[\sqrt{80 y} = \sqrt{16 \cdot 5 y} = 4 \sqrt{5 y}.\]
Substituting this back into the second term, we get:
\[3 \cdot 4 \sqrt{5 y} = 12 \sqrt{5 y}.\]
Combining these parts, the simplified expression is \(6 y^2 \sqrt{10} + 12 \sqrt{5 y}\). Given the provided choices, there appears to be a typo in the options as none exactly match this result; however, based on standard simplification procedures, this is the definitive algebraic equivalent.
Examples
Understanding how to simplify radical expressions is fundamental in fields like engineering and physics. For example, when calculating the energy or resistance in a system where variables are under square roots, simplifying expressions helps in making the formulas manageable for further analysis and design. Mastering these techniques ensures that complex physical laws can be represented in their most concise and usable form.