Problem 60 - Algebra for College Students

Question

Write each of the following in terms of \(i\), perform the indicated operations, and simplify. For example, $$ \begin{aligned} \sqrt{-3} \sqrt{-8} &=(i \sqrt{3})(i \sqrt{8}) \\ &=i^{2} \sqrt{24} \\ &=(-1) \sqrt{4} \sqrt{6} \\ &=-2 \sqrt{6} \end{aligned} $$ $$\frac{\sqrt{-96}}{\sqrt{2}}$$

Step-by-Step Solution

Answer

\(4i\sqrt{3}\)

Step 1: Express in terms of i

Start with the expression \(\frac{\sqrt{-96}}{\sqrt{2}}\). The negative inside the square root indicates you need to express it in terms of \(i\). This means \(\sqrt{-96} = i\sqrt{96}\).

Step 2: Simplify the Radical

Simplify \(\sqrt{96}\) by finding its prime factors: \(96 = 2^5 \cdot 3\). Therefore, \(\sqrt{96} = \sqrt{(2^4) \cdot 2 \cdot 3}\).

Step 3: Break Down the Expression

Use the property of square roots \(\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}\). So, \(\sqrt{96} = \sqrt{2^4} \cdot \sqrt{2 \cdot 3} = 4\sqrt{6}\).

Step 4: Substitute and Divide

Substitute \(\sqrt{96} = 4\sqrt{6}\) back into the expression and manage the \(i\): \(\frac{i \cdot 4\sqrt{6}}{\sqrt{2}} = i \cdot \frac{4\sqrt{6}}{\sqrt{2}}\).

Step 5: Simplify the Division

Further simplify by dividing the expression inside the square root: \(\frac{\sqrt{6}}{\sqrt{2}} = \sqrt{\frac{6}{2}} = \sqrt{3}\).

Step 6: Final Simplification

Complete the calculation: \(4 \cdot \sqrt{3} = 4\sqrt{3}\). Therefore, the final answer is \(4i\sqrt{3}\).

Key Concepts

Imaginary Unit

The imaginary unit, often denoted as \(i\), plays a crucial role in handling square roots of negative numbers. In standard mathematical terms, the imaginary unit is defined as \(i = \sqrt{-1}\). You'll come across \(i\) routinely when dealing with complex numbers, which are numbers that have a real part and an imaginary part.

Understanding \(i\) is fundamental because whenever you encounter the square root of a negative number, converting it into \(i\) makes calculations more manageable. The statement \(\sqrt{-x} = i\sqrt{x}\) lets us work with these imaginary components effectively.

Some important properties of \(i\) include:

• \(i^2 = -1\)
• \(i^3 = -i\)
• \(i^4 = 1\)

These properties cycle every four powers and are essential when performing operations with complex numbers, especially when you're simplifying expressions that involve \(i\).

Radical Simplification

Radical simplification involves breaking down square roots into their simplest form. Often, this means factoring a number into its prime components and simplifying. Let's delve into the example of \(\sqrt{96}\).

Start by finding the prime factorization of 96. We get that \(96 = 2^5 \times 3\). To simplify \(\sqrt{96}\), apply square root properties:

• \(\sqrt{96} = \sqrt{2^5 \times 3}\).
• Break it into \(\sqrt{(2^4) \times 2 \times 3}\), where \(2^4\) forms a perfect square.
• Since \(\sqrt{2^4} = 4\), you can simplify this to \(4\sqrt{6}\).

By factoring numbers into primes, you untangle a square root into simpler parts. This principle is the backbone for simplifying complex radicals and makes it easier to perform further arithmetic operations such as multiplication or division.

Square Roots

Square roots are essential in mathematics, providing the means to find a number which, when multiplied by itself, gives the original number. For instance, the square root of 9 is 3 because \(3^2 = 9\).

When handling square roots in expressions, especially those with negative radicands like \(\sqrt{-96}\), it's critical to recognize they transform using the imaginary unit \(i\). This transforms our negative square root into \(i\sqrt{96}\).

Rules and properties you must remember include:

• \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\)
• \(\sqrt{a/b} =\frac{\sqrt{a}}{\sqrt{b}}\)

These assist in simplifying expressions, allowing smoother manipulation.

Square roots encourage a keen observation of numbers to identify opportunities for simplification, aiding tremendously in computations related to complex numbers. This understanding comes in handy for exercises where you're instructed to simplify and express radicals, as shown in the original problem set.