Problem 39 - Algebra 2

Question

Solve. Check for extraneous solutions. \((2 x+3)^{\frac{3}{4}}-3=5\)

Step-by-Step Solution

Answer

The solution to the equation \((2x+3)^{\frac{3}{4}}-3=5\) is \(x=14.5\).

Step 1: Isolate the Radical

First, isolate the radical on one side of the equation. This gives:\[(2 x+3)^{\frac{3}{4}} = 5+3=8.\]

Step 2: Remove the Fraction Exponent

Raise both sides to the power of \(\frac{4}{3}\) to remove the fraction exponent in the equation. This gives:\[((2 x+3)^{\frac{3}{4}})^{\frac{4}{3}} = 8^{\frac{4}{3}}\(2 x+3) = 8^{\frac{4}{3}}.\]

Step 3: Calculate the Expression on the Right

Calculate the expression on the right side of the equation : \[(2 x+3) = 8^{\frac{4}{3}} = 32.\]

Step 4: Solve for x

Solve the equation for x:\(2 x= 32-3 = 29)\x = \frac{29}{2}=14.5.\]

Step 5: Check for Extraneous Solutions

Substitute \(x=14.5\) into the original equation to ensure it is not an extraneous solution:\(\((2 (14.5)+3)^{\frac{3}{4}} = 8), which simplifies to \(8^{\frac{3}{4}} = 8\), indicating that \(x=14.5\) is a valid solution.

Key Concepts

Fractional Exponents

Fractional exponents are a way to express power and roots together in a single notation. This representation can simplify complex mathematical operations. A fractional exponent like \(a^{m/n}\) means that you first raise \(a\) to the power of \(m\), and then take the \(n\)-th root of the result.

• The numerator (\(m\)) is the power to be applied.
• The denominator (\(n\)) represents the root to be taken.

When solving equations involving fractional exponents, it is often easier to deal with them by isolating the exponent part on one side. You can then eliminate the fractional exponent by raising both sides of the equation to a power that cancels the fraction, which was demonstrated by raising the equation to the power of \(\frac{4}{3}\) in the original exercise.
This clears out the exponent and simplifies solving for the unknown variable. Understanding this concept is crucial because it allows you to handle equations where variables are within roots and powers, which is common in algebra.

Extraneous Solutions

Extraneous solutions are solutions that emerge as a result of the process of solving the equation, but do not actually satisfy the original equation. They often occur in radical and rational equations, where the steps involved in manipulating the equation can introduce solutions that weren't originally true.
To identify extraneous solutions:

• Always substitute the solution back into the original equation.
• Check if both sides of the equation are equal after the solution is substituted.

In the exercise provided, checking for extraneous solutions involved substituting \(x = 14.5\) back into the original equation, ensuring that both sides resulted in the same value and confirming that \(x = 14.5\) was not extraneous. This step is important because it validates that the solution obtained is indeed correct and relevant to the initial problem.

Solving Equations Step-by-Step

Approaching problems step-by-step helps break down complex equations into more manageable parts. This methodical approach involves several crucial stages:

• Isolate Terms: Start by moving terms to one side to isolate the radical or power terms. For instance, isolate \((2x+3)^{\frac{3}{4}}\) by moving all other terms to the opposite side.
• Eliminate Exponents: Use appropriate mathematical operations that counteract the powers, such as raising both sides to an inversely calculated power.
• Solve for the Variable: Once the exponents are removed, solve for the variable using basic algebraic operations like addition, subtraction, or division.
• Check Solutions: Finally, substitute the variable back into the initial equation to ensure it provides a true statement.

This methodical process ensures clarity and increases accuracy, allowing one to effectively find solutions to radical and complex equations consistently. By practicing this approach, students can build confidence in navigating through intricate algebraic problems.