- Simplify the first term: \(2 \sqrt{25 y^8} = 2 \cdot 5 \cdot y^4 = 10 y^4\).
- Simplify the second term: \(4 \sqrt{6 y^3} = 4 \cdot y \cdot \sqrt{6y} = 4y \sqrt{6y}\).
- Multiply the simplified terms: \(10 y^4 \cdot 4 y \sqrt{6y} = 40 y^5 \sqrt{6y}\).
- Match the result to the second option: \(\boxed{40 y^5 \sqrt{6 y}}\).
Explanation
- Analysis of the Expression
We are given the expression \(2 \sqrt{25 y^8} \cdot 4 \sqrt{6 y^3}\) with the condition \(y > 0\). To find the equivalent expression, we will simplify each radical part separately and then multiply them together using the properties of exponents and square roots.
- Simplifying the First Term
First, let's simplify the first term \(2 \sqrt{25 y^8}\).
Since \(25\) is a perfect square, \(\sqrt{25} = 5\).
For the variable part, using the property \(\sqrt{y^n} = y^{n/2}\) for \(y > 0\), we have \(\sqrt{y^8} = y^{8/2} = y^4\).
Thus, \(2 \sqrt{25 y^8} = 2 \cdot 5 \cdot y^4 = 10 y^4\).
- Simplifying the Second Term
Next, let's simplify the second term \(4 \sqrt{6 y^3}\).
We can rewrite \(\sqrt{y^3}\) as \(\sqrt{y^2 \cdot y}\). Since \(y > 0\), \(\sqrt{y^2} = y\), so \(\sqrt{y^3} = y \sqrt{y}\).
Substituting this back into the term, we get \(4 \sqrt{6 y^3} = 4 \sqrt{6} \cdot y \sqrt{y} = 4 y \sqrt{6y}\).
- Combining and Selecting the Correct Answer
Now, we multiply the two simplified expressions:
\((10 y^4) \cdot (4 y \sqrt{6y}) = (10 \cdot 4) \cdot (y^4 \cdot y) \cdot \sqrt{6y} = 40 y^5 \sqrt{6y}\).
Comparing this result to the given options:
- \(14 y^5 \sqrt{6 y^2}\)
- \(40 y^5 \sqrt{6 y}\)
- \(17 y^2 \sqrt{6 y^5}\)
- \(8 y^2 \sqrt{150 y}\)
The expression \(40 y^5 \sqrt{6y}\) matches the second option.
Examples
In engineering and physics, when calculating power dissipation in a circuit where the resistance depends on a variable \(y\) representing time or temperature, you often encounter products of radical expressions. Simplifying these expressions allows for easier computation and analysis of the system's behavior over time. For example, if you have a component whose amplitude is determined by \(2\sqrt{25y^8}\) and another factor of \(4\sqrt{6y^3}\), the combined effect is best represented by the simplified product \(40y^5\sqrt{6y}\), which clearly shows the power-law relationship between the total output and the variable \(y\).