Problem 8 - College Algebra

Question

For the following exercises, solve the rational exponent equation. Use factoring where necessary. $$ 2 x^{\frac{1}{2}}-x^{\frac{1}{4}}=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = \frac{1}{16}\).

Step 1: Identify Common Terms

Notice the terms in the equation: \(2x^{\frac{1}{2}}\) and \(x^{\frac{1}{4}}\). The exponents are fractions, which are called rational exponents.

Step 2: Substitute Variables

To make the equation easier to work with, let's set \(y = x^{\frac{1}{4}}\). Then, \(y^2 = x^{\frac{1}{2}}\). This changes the equation to:\[2y^2 - y = 0\]

Step 3: Factoring

The equation \(2y^2 - y = 0\) can be factored by taking \(y\) common:\[y(2y - 1) = 0\]

Step 4: Solve Each Factor

From \(y(2y - 1) = 0\), set each factor to zero: 1. \(y = 0\)2. \(2y - 1 = 0\), which simplifies to \(y = \frac{1}{2}\).

Step 5: Substitute Back to Original Variable

Remember \(y = x^{\frac{1}{4}}\). Substitute back:1. If \(y = 0\), then \(x^{\frac{1}{4}} = 0\), which implies \(x = 0\).2. If \(y = \frac{1}{2}\), then \(x^{\frac{1}{4}} = \frac{1}{2}\). To isolate \(x\), raise both sides to the power of 4:\(x = \left(\frac{1}{2}\right)^4 = \frac{1}{16}\).

Step 6: Conclusion

The solutions to the equation are \(x = 0\) and \(x = \frac{1}{16}\).

Key Concepts

Factoring

Factoring is a mathematical process where you express an expression as a product of its factors. This process can simplify equations, making them easier to solve. In the case of polynomial equations, like the transformed version of our exercise, factoring helps isolate solutions we need. Let's break down the steps involved:

• Identify common terms: Look for terms that repeat or can be grouped together. This often involves common variables or coefficients.
• Use distributive properties: For expressions like \(2y^2 - y\), factor out common variables or numbers. This particular expression factors to \(y(2y - 1) = 0\).
• Solve each factor: Each factor is set to zero, giving simpler equations to solve. For example, \(y = 0\) and \(2y - 1 = 0\) represent two different parts of the solution.

The method effectively reduces a quadratic equation into simple linear solutions, which are much easier to manage.

Substitution Method

The substitution method is a tactical approach to simplify complex equations by introducing a new variable. In equations with rational exponents, substitution can turn an intimidating equation into a more familiar form:

• Replace complex terms: For the given equation \(2x^{\frac{1}{2}} - x^{\frac{1}{4}} = 0\), replacing \(x^{\frac{1}{4}}\) with \(y\) helps make the equation easier to handle.
• Transform the equation: With \(y = x^{\frac{1}{4}}\), it follows that \(y^2 = x^{\frac{1}{2}}\). This leads to a simpler polynomial: \(2y^2 - y = 0\).
• Solve using the new variable: This form can then be solved using factorization or other simple methods.

Substitution is particularly useful when dealing with rational exponents, as it temporarily removes them, making the equation more straightforward to solve.

Solving Equations

Solving equations involves finding the values of the variables that make the equation true. The process usually entails a few essential steps, especially when dealing with rational exponents and the use of factoring:

• Isolate variable expressions: Begin by isolating terms involving the variable using algebraic manipulation, such as moving like terms to one side.
• Use procedures like factoring or substitution to simplify: Either solve directly or simplify the equation using methods like factoring or substitution to break down complex expressions.
• Solve each simplified equation: After simplifying, equations should become straightforward to solve. In our exercise, once factored, we solve \(y(2y - 1) = 0\) resulting in \(y = 0\) or \(y = \frac{1}{2}\).
• Revert any substitution: Finally, if substitution was used, revert it back to the original variable to find all possible solutions.

For our particular problem, solving resulted in \(x = 0\) and \(x = \frac{1}{16}\). Each solution directly addresses the original equation and validates the method used.