Problem 69 - Algebra and Trigonometry

Question

\(47-70\) The given equation involves a power of the variable. Find all real solutions of the equation. $$ 2 x^{5 / 3}+64=0 $$

Step-by-Step Solution

Answer

The real solution is \(x = -8\).

Step 1: Rearrange the Equation

Begin by moving constants from one side of the equation to the other. Subtract 64 from both sides to isolate the term with the variable on one side:\[ 2x^{5/3} = -64 \]

Step 2: Isolate the Variable Term

Divide both sides of the equation by 2 to further isolate the term with the variable:\[ x^{5/3} = -32 \]

Step 3: Eliminate the Fractional Exponent

Raise both sides of the equation to the power of \(3/5\) in order to cancel the fractional exponent on \(x\):\[ x = (-32)^{3/5} \]

Step 4: Calculate the Root and the Power

First, calculate the fifth root of \(-32\). The fifth root of \(-32\) is \(-2\) because \((-2)^5 = -32\). Then raise \(-2\) to the third power:\[ x = (-2)^3 = -8 \]

Key Concepts

Fractional Exponents

In algebra, we often encounter exponents in fractional form, such as in the equation \(2x^{5/3} + 64 = 0\). Fractional exponents offer a handy way to express roots and powers compactly. Here’s how it works:

• The numerator of a fractional exponent indicates the power to which a number is raised.
• The denominator gives you the root you need to take.

For example, in \(x^{5/3}\), the numerator \(5\) denotes the fifth power, and the denominator \(3\) denotes the cube (or third) root. It is essential to understand how to handle these exponents as they allow complex equations to be simplified effectively.
Fractional exponents enable us to seamlessly switch between exponentiation and root extraction, a vital skill when solving higher-level algebraic equations.

Isolation of Variables

When solving an algebraic equation, the goal is often to isolate the variable. This means getting the variable all by itself on one side of the equation to find its value. Let's see how this concept was applied in our example:

• First, move all non-variable terms to the opposite side of the equation. In \(2x^{5/3} + 64 = 0\), subtract 64 to isolate the term with \(x^{5/3}\): \(2x^{5/3} = -64\).
• Next, simplify further by dividing the equation by 2 to completely isolate the variable term: \(x^{5/3} = -32\).

By isolating the variable, you streamline solving the equation, making it easier to focus on eliminating the remaining complexities, like the fractional exponent.

Roots and Powers

Roots and powers are fundamental operations in algebra. In our example, once the variable term is isolated, we tackle it by dealing with roots and powers:

• When dealing with a term such as \(x^{5/3}\), you can eliminate the fractional exponent through root extraction and raising to a power. Raise both sides to the power of \(3/5\), reversing the initial operation: \(x = (-32)^{3/5}\).
• Calculate the fifth root first, as per the denominator: the fifth root of \(-32\) is \(-2\) since \((-2)^5 = -32\).
• Then raise \(-2\) to the third power: \((-2)^3 = -8\), which is the solution to the equation.

Understanding how to effectively maneuver between roots and powers not only helps in solving equations but also in simplifying and transforming expressions in various mathematical contexts.