Problem 22 - Intermediate Algebra

Question

Use radical notation to write each expression. Simplify if possible. $$ (-9)^{3 / 2} $$

Step-by-Step Solution

Answer

\((-9)^{3/2} = -27i\).

Step 1: Convert to Radical Notation

To express \((-9)^{3/2}\) using radical notation, we recognize that the exponent \(\frac{3}{2}\) can be split into two parts: the denominator 2 and the numerator 3. The denominator 2 indicates a square root, and the numerator 3 implies cubing. The expression becomes \((-9)^{3/2} = \left(\sqrt{-9}\right)^3\).

Step 2: Simplify the Square Root

Next, simplify \(\sqrt{-9}\). Since we are dealing with a negative under a square root, we use imaginary numbers. \(\sqrt{-9}\) can be simplified to \(3i\), where \(i = \sqrt{-1}\). Therefore, \(\left(\sqrt{-9}\right)^3 = (3i)^3\).

Step 3: Cube the Result

Now we need to cube \(3i\). Use the formula for cubing: \((a + bi)^3 = a^3 + 3a^2bi - 3ab^2 - b^3i\) and remember that \(i^2 = -1\). For our expression \((3i)^3\), cube each part: \(3^3i^3 = 27i^3\). Simplifying \(i^3\) gives \(i^3 = i^2 \cdot i = -1 \cdot i = -i\). Thus, \(27i^3 = -27i\).

Key Concepts

Imaginary Numbers

Imaginary numbers can initially seem a bit puzzling! But they're not as complex as they sound. An imaginary number is the square root of a negative number. For instance, when you see \(-9\) under a square root, direct calculation isn't possible using real numbers. Instead, we use the imaginary unit, which we call \(i\). \(i\) is defined as \(\sqrt{-1}\), and every imaginary number is a multiple of this unit.
So, if you have \(\sqrt{-9}\), you rewrite it using \(i\), like this: \(\sqrt{-9} = \sqrt{9 \times -1} = \sqrt{9}\cdot \sqrt{-1} = 3i\).
This forms a basic imaginary number, expressing a concept that isn’t "real" in the traditional sense but is very useful in complex calculations.

• Key idea: Imaginary numbers use the fact that \(i^2 = -1\).
• Notation: Imaginary numbers are expressed in terms of \(i\).

Square Root

The square root of a number is essentially a value that, when multiplied by itself, results in that original number. For example, the square root of 9 is 3 because \(3 \times 3 = 9\). But what if the number is negative?
With negative numbers under the square root, you dip into the realm of imaginary numbers. That's why \(\sqrt{-9} = 3i\), wherein \(i\) accounts for the negative part.
Square roots are denoted either as \(\sqrt{x}\) or by an exponent of \(\frac{1}{2}\). Let’s remember:

• Positive \(x\): Simple root, like \(\sqrt{9} = 3\).
• Negative \(x\): Requires imaginary unit, thus \(\sqrt{-9} = 3i\).

This wraps up square roots: DMagnificent for uncovering hidden roots of a number and experimenting with imaginary numbers too!

Exponents

Exponents are like mathematical shorthand for repeated multiplication. When you see a number raised to a power, you're dealing with an exponent. For example, \(2^3\) means \(2 \times 2 \times 2\), which equals 8.
In the given expression, \((-9)^{3/2}\), it signifies both taking a square root (because of the denominator 2) and raising to the power 3 (from the numerator). Hence, when we work with fractional exponents, we often break them into two steps: the root step and the power step.

• Fractional exponent = root + power: \((-9)^{3/2} = \left(\sqrt{-9}\right)^3\).
• Whole number exponents: Multiply the base repeatedly.

Utilizing exponents streamlines complex calculations, especially when mixed with roots or imaginary numbers!

Simplifying Expressions

Simplifying mathematical expressions means breaking them down into their simplest form. It's like cleaning up clutter to see the core idea.
Sometimes, expressions include crazy combinations of roots, powers, and even imaginary numbers. The key is to simplify step by step. Look at the original problem \((-9)^{3/2}\). After converting it to radical notation and simplifying the root to produce imaginary numbers, you then need to cube the outcome.
Consider this process:

• Convert: Turn complex fractions into roots and powers.
• Simplify: Handle roots and imaginary units step-by-step.
• Cubic Function: When cubing \(3i\), you use exponent rules.

Finally, the expression simplifies to \(-27i\), beautifully neat and free of unnecessary complexity!