Problem 40 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solving an Equation Involving Rational Exponents Find all solutions of the equation algebraically. Check your solutions. $$(x-1)^{3 / 2}=8$$

Step-by-Step Solution

Answer

The equation \((x-1)^{3 / 2}=8\) does not have a valid solution.

Step 1: Remove the Rational Exponent

To take away the rational exponent and simplify the problem, cube root both sides of the equation: \[\sqrt[3]{(x-1)^3} = \sqrt[3]{8}\]This simplifies to:\[x-1=2\]

Step 2: Solve for x

To find \(x\), add 1 to both sides of the equation.\(x-1+1=2+1\)This simplifies to:\(x=3\)

Step 3: Check the Solution

To ensure that \(x=3\) is a valid solution and not an extraneous solution, substitute \(x=3\) into the original equation: \[(3-1)^{3 / 2} = 8\]This simplifies to \(2^{3/2} = \sqrt[3]{8}\), which is not valid. Hence, \(x=3\) is not a valid solution.

Key Concepts

Understanding Rational Exponents

When we come across expressions with rational exponents in algebra, it's important to understand how to interpret and work with them. Rational exponents like \( \frac{m}{n} \) are another way of expressing roots, specifically the n-th root of a number raised to the m-th power. The equation \( (x-1)^{\frac{3}{2}} = 8 \) involves a rational exponent, which combines a square root (because of the denominator 2) and a cube (because of the numerator 3).

To solve such equations, we often remove the rational exponent by rewriting the expression with a radical. In our case, taking the cube root is equivalent to raising each side of the equation to the power of \( \frac{2}{3} \) (because \( \frac{2}{3} \) is the reciprocal of \( \frac{3}{2} \) ). Through this process, we're left with a simpler, equivalent algebraic expression that can be solved more straightforwardly.

Solving Algebraic Equations

Algebraic equations are mathematical statements that assert the equality of two expressions. They generally involve variables, constants, and arithmetic operations. The process of finding the value(s) of the variable(s) that make the equation true is known as solving the equation.

In the context of our exercise with the equation \( (x-1)^{\frac{3}{2}} = 8 \) we approach solving it systematically. Firstly, we remove the rational exponent which simplifies our problem to \( x-1 = 2 \) after taking the cube root of both sides. Then, we solve for \( x \) by isolating the variable, which requires us to perform inverse operations. In this case, we added 1 to both sides to effectively ‘undo’ the subtraction of 1 from \( x \) which resulted in finding the value of \( x \) to be 3.

Addressing Extraneous Solutions

Extraneous solutions are results that emerge from the process of solving equations but do not actually satisfy the original equation. They often appear when we perform operations such as squaring both sides of an equation or taking a root, which can introduce additional solutions. That is why checking solutions in the context of the original equation is a crucial step to validate their correctness.

Using our example \( (x-1)^{\frac{3}{2}} = 8 \) and its apparent solution \( x = 3 \), when we substitute back into the original equation, we find that it does not hold true. Thus, \( x = 3 \) turns out to be an extraneous solution, and we have to discard it. This emphasizes the importance of always verifying each potential solution by plugging it back into the original equation to ensure it truly works, preventing any misinterpretation. Checking solutions not only confirms their validity but also enhances our understanding of the characteristics of the equations we are working with.