Problem 33 - Elementary and Intermediate Algebra

Question

Simplify each expression. Assume that the variables can be any real number, and use absolute value symbols when necessary. See Example 2 . $$ \left(-64 p^{8}\right)^{1 / 2} $$

Step-by-Step Solution

Answer

The simplified expression is \(8i p^4\).

Step 1: Apply the Power of a Product Property

The expression is \( \left(-64 p^8\right)^{1/2} \). First, break down the expression inside the parentheses and apply the exponent to each factor separately. Thus, it becomes \((-64)^{1/2} \times (p^8)^{1/2}\).

Step 2: Simplify the Numerical Expression

Calculate \((-64)^{1/2}\). The square root of \(-64\) isn't a real number since square roots of negative numbers are imaginary. Because we're simplifying in terms of real numbers, let's focus on extracting the real part, \(64^{1/2}\), and note the need for the imaginary unit \(i\). So \((-64)^{1/2} = 8i\).

Step 3: Simplify the Algebraic Expression

Now, consider \((p^8)^{1/2}\). Using the rule for powers, this simplifies to \(p^{8 \times \frac{1}{2}} = p^4\).

Step 4: Combine Simplified Parts

Combine the results from Steps 2 and 3. The expression becomes \(8i \times p^4\). This is the simplified form of the original expression given that it involves the imaginary unit \(i\).

Key Concepts

Power of a Product Property

When faced with expressions like \((-64 p^8)^{1/2}\), the Power of a Product Property is a handy tool to break things down. This property allows us to distribute an exponent over a product, applying the exponent to each factor individually.
For example, with our task at hand:

• Start by looking at the expression inside the parenthesis: \(-64 p^8\).
• According to the Power of a Product Property, apply the exponent \(\frac{1}{2}\) to each part separately: \((-64)^{1/2} \times (p^8)^{1/2}\).

By distributing the exponents, our job becomes simpler. This is because you can now focus on solving smaller, less complex pieces.
In mathematics, breaking down big tasks into small parts is sometimes half the victory!

Imaginary Numbers

You might wonder about that minus sign inside our square root, \((-64)^{1/2}\). That's where imaginary numbers step in. A number that when squared gives a negative result is imaginary.

Here's how it works:

• The negative sign under the root tells us we're dealing with a number that isn’t real in our typical number sense.
• Imaginary numbers are represented using the symbol \(i\), where \(i\) is defined as the square root of \(-1\).

Therefore, the square root calculations proceed like so:

• Inside the square root, separate the negative sign, so basically we're finding \(\sqrt{64} \cdot \sqrt{-1}\).
• This translates to \(8 \cdot i\), which simplifies \((-64)^{1/2}\) to \(8i\).

This is how imaginary numbers can transform a seemingly unsolvable calculation into something solvable! It’s like having a mathematical wildcard to overcome the hurdle of negative square roots.

Exponents in Algebra

Working with exponents can seem like an intimidating task at first. In algebra, exponents signify repeated multiplication. The expression \(p^8\) means multiplying \(p\) by itself eight times.

Here's how to simplify expressions like \((p^8)^{1/2}\):

• When an exponent is raised to a power, multiply the exponents together. That's the rule at play here: \((x^a)^b = x^{a \times b}\).
• So in our example, \((p^8)^{1/2} = p^{8 \times \frac{1}{2}} = p^4\).

This rule shows that simplifying powers can truly slice down the complexity. You emerge with a more approachable expression, in this case, \(p^4\).
Understanding how these rules apply makes algebraic life much more manageable, allowing you to simplify and conquer daunting expressions with greater ease!