- Evaluate arithmetic radicals by direct computation: \(\sqrt{16-12} = 2\) and \(\sqrt{81+360} = 21\).
- Simplify expression 14 using exponent properties: \(x^{21/3} \cdot x^{13/2} \cdot x^{1/2} = x^{14}\).
- Simplify expression 15 by calculating the root: \(-y^{15/5} = -y^3\).
- Simplify expression 16 by applying the power rule: \(4(3x)^3 = 108x^3\).
- Final results: 12. \(\boxed{2}\), 13. \(\boxed{21}\), 14. \(\boxed{x^{14}}\), 15. \(\boxed{-y^3}\), 16. \(\boxed{108x^3}\).
Explanation
- Problem Analysis
We need to evaluate five radical expressions: 12, 13, 14, 15, and 16. For problems 12 and 13, we perform arithmetic calculations. For problems 14, 15, and 16, we simplify expressions involving variables, keeping in mind that all variables represent positive real numbers. We will use the properties of exponents and radicals: \(\sqrt[n]{x^m} = x^{m/n}\), \((x^a)^b = x^{ab}\), and \(x^a \cdot x^b = x^{a+b}\).
- Evaluating Arithmetic Radicals
For expression 12: \(\sqrt{4^2 - 4(3)(1)} = \sqrt{16 - 12} = \sqrt{4} = 2\).
For expression 13: \(\sqrt{9^2 - 4(10)(-9)} = \sqrt{81 - (-360)} = \sqrt{81 + 360} = \sqrt{441} = 21\).
- Simplifying Algebraic Radicals
For expression 14: \(\sqrt[3]{x^{21}} \cdot \sqrt{x^3 \cdot x^{10}} \cdot \sqrt{x} = x^{21/3} \cdot \sqrt{x^{13}} \cdot x^{1/2} = x^7 \cdot x^{13/2} \cdot x^{1/2} = x^7 \cdot x^{(13/2 + 1/2)} = x^7 \cdot x^{14/2} = x^7 \cdot x^7 = x^{14}\).
For expression 15: \(-\sqrt[5]{y^{15}} = -(y^{15/5}) = -y^3\).
For expression 16: \(4\sqrt{(3x)^6} = 4(3x)^{6/2} = 4(3x)^3 = 4(27x^3) = 108x^3\).
- Final Answers
The evaluation results are as follows:
Problem 12: \(\boxed{2}\)
Problem 13: \(\boxed{21}\)
Problem 14: \(\boxed{x^{14}}\)
Problem 15: \(\boxed{-y^3}\)
Problem 16: \(\boxed{108x^3}\)
Examples
Simplifying radical expressions is a foundational skill in fields like engineering and physics. For example, when calculating the magnitude of a vector or finding the period of a pendulum, you frequently encounter square and cube roots. Being able to simplify complex radical terms into standard power forms allows for much easier arithmetic and algebraic manipulation when solving engineering design equations or analyzing physical systems.