- Express \(\sqrt{-x}\) as \(i\sqrt{x}\) to simplify both expressions.
- For the first expression, \(\frac{i\sqrt{70}}{i\sqrt{10}}\), cancel \(i\) and divide under the radical to get \(\sqrt{7}\).
- For the second expression, \((i\sqrt{10})(i\sqrt{2}) = i^2\sqrt{20}\), simplify to \(-1 \cdot 2\sqrt{5} = -2\sqrt{5}\).
- The simplified answers are \(\boxed{\sqrt{7}}\) and \(\boxed{-2\sqrt{5}}\).
Explanation
- Analysis and Definitions
We are given two expressions to simplify: \(\frac{\sqrt{-70}}{\sqrt{-10}}\) and \(\sqrt{-10} \cdot \sqrt{-2}\). We recall that for any positive real number \(x\), the square root of a negative number is defined as \(\sqrt{-x} = i\sqrt{x}\), where \(i = \sqrt{-1}\) is the imaginary unit. Our goal is to perform these simplifications step-by-step while ensuring no negative numbers remain under radicals and the final forms are as simplified as possible.
- Simplifying the First Expression
For the first expression, \(\frac{\sqrt{-70}}{\sqrt{-10}}\), we use the property \(\sqrt{-x} = i\sqrt{x}\) to rewrite both the numerator and the denominator:
\[\frac{\sqrt{-70}}{\sqrt{-10}} = \frac{i\sqrt{70}}{i\sqrt{10}}.\]
We can cancel the imaginary unit \(i\) from both the numerator and the denominator:
\[\frac{i\sqrt{70}}{i\sqrt{10}} = \frac{\sqrt{70}}{\sqrt{10}}.\]
Using the property of radicals \(\frac{\sqrt{a}}{\sqrt{b}} = \sqrt{\frac{a}{b}}\) for \(a, b > 0\), we get:
\[\sqrt{\frac{70}{10}} = \sqrt{7}.\]
Since \(\sqrt{7}\) is already in its simplest form, the first expression simplifies to \(\sqrt{7}\).
- Simplifying the Second Expression
For the second expression, \(\sqrt{-10} \cdot \sqrt{-2}\), we again rewrite each radical using \(i\):
\[\sqrt{-10} \cdot \sqrt{-2} = (i\sqrt{10}) \cdot (i\sqrt{2}).\]
Grouping the terms, we have:
\[(i \cdot i) \cdot (\sqrt{10} \cdot \sqrt{2}) = i^2 \cdot \sqrt{10 \cdot 2} = i^2 \cdot \sqrt{20}.\]
Recalling that \(i^2 = -1\), the expression becomes:
\[-1 \cdot \sqrt{20} = -\sqrt{20}.\]
We simplify the radical \(\sqrt{20}\) by noting that \(20 = 4 \cdot 5\), so \(\sqrt{20} = \sqrt{4 \cdot 5} = 2\sqrt{5}\). Thus, the expression becomes:
\[-\sqrt{20} = -2\sqrt{5}.\]
The second expression simplifies to \(-2\sqrt{5}\).
- Conclusion and Final Results
Based on the calculations above, we have simplified both expressions:
- \(\frac{\sqrt{-70}}{\sqrt{-10}} = \sqrt{7}\)
- \(\sqrt{-10} \cdot \sqrt{-2} = -2\sqrt{5}\)
Thus, the final results are \(\boxed{\sqrt{7}}\) and \(\boxed{-2\sqrt{5}}\).
Examples
Understanding complex numbers and their arithmetic is crucial in fields like electrical engineering, where imaginary numbers are used to represent phase shifts in alternating current circuits. For instance, when analyzing impedance in a circuit with capacitors and inductors, you often encounter square roots of negative values. Mastering these manipulations allows engineers to model and solve complex circuit behaviors efficiently.