Problem 44 - Algebra for College Students

Question

Write each of the following in terms of \(i\), perform the indicated operations, and simplify. For example, $$ \begin{aligned} \sqrt{-3} \sqrt{-8} &=(i \sqrt{3})(i \sqrt{8}) \\ &=i^{2} \sqrt{24} \\ &=(-1) \sqrt{4} \sqrt{6} \\ &=-2 \sqrt{6} \end{aligned} $$ $$\sqrt{-81} \sqrt{-25}$$

Step-by-Step Solution

Answer

The simplified result is \(-45\).

Step 1: Express terms in terms of i

To express each square root in terms of \(i\), recall that \(i = \sqrt{-1}\). Thus, \(\sqrt{-81} = \sqrt{81} \cdot \sqrt{-1} = 9i\) and \(\sqrt{-25} = \sqrt{25} \cdot \sqrt{-1} = 5i\).

Step 2: Multiply the terms

Multiply the two expressions with \(i\): \((9i)(5i)\).

Step 3: Simplify the multiplication

Apply the multiplication of \(i\) terms: \((9 \cdot 5)(i \cdot i) = 45i^2\).

Step 4: Substitute and simplify \(i^2\)

By definition, \(i^2 = -1\), so replace \(i^2\) with -1 in the equation: \(45i^2 = 45(-1) = -45\).

Key Concepts

The Imaginary Unit

When we talk about complex numbers, we often mention something called the imaginary unit, denoted as \(i\). This nifty little symbol is defined as \(i = \sqrt{-1}\). Why does this matter? Well, because without it, solving certain equations would be impossible! We use \(i\) to handle square roots of negative numbers neatly.

Let's say we're faced with \(\sqrt{-9}\). This can be tricky, but with \(i\), we can rewrite it as \(3i\) because \(\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i\). Pretty cool, right?

Here are a few quick reminders about \(i\):

• \(i^2 = -1\), making it unique.
• It can be used to simplify complex roots.
• It's a fundamental part of complex number operations.

Multiplication of Complex Numbers

Multiplying complex numbers might sound tricky at first, but it's very similar to multiplying regular numbers, with the bonus of dealing with \(i\). When two complex numbers are multiplied, say \((a + bi)\) and \((c + di)\), you can employ the distributive property to get, \((a + bi)(c + di) = ac + adi + bci + bdi^2\).

After this, you group like terms and remember that \(i^2 = -1\), which helps to combine real and imaginary parts correctly. For example, when you multiply \((9i)\) and \((5i)\), multiply the numbers first like \(9 \times 5 = 45\), and then do the same with \(i\) giving \(i^2\). Combine it to get \(45i^2\).

Simplifying \(45i^2\) with \(i^2 = -1\), it turns into \(-45\). So, from a multiplication involving imaginary components, we ended up with a real number.

Simplifying Radicals

Handling squared numbers is straightforward until negatives come into play. That’s when we turn to simplifying radicals to make life easier! When you encounter a radical like \(\sqrt{-81}\), the imaginary unit \(i\) becomes your friend. You separate the radical into two parts: the positive square root and the square root of \(-1\).

For \(\sqrt{-81}\), rewrite it as \(\sqrt{81} \cdot \sqrt{-1}\). So, we determine \(\sqrt{81} = 9\) and then apply our trusty \(i\) to represent \(\sqrt{-1}\), concluding with \(9i\).

To simplify further, consider multiplying radicals. This follows a similar pattern. Suppose you have \(\sqrt{-81} \cdot \sqrt{-25}\), transform each radical: \(9i\) and \(5i\), multiply them together, using rules of multiplication of complex numbers to end up with something more manageable.

• This often results in a completely real number.
• Use \(i^2 = -1\) to ensure any imaginary components are neatly resolved.

Simplifying these radicals ensures everything stays organized and interpretable. It’s all about transforming the complex into clarity!