Problem 31 - Precalculus: Functions and Graphs

Question

Solve the equation. \(y^{3 / 2}=5 y\)

Step-by-Step Solution

Answer

The solutions are \( y = 0 \) and \( y = 25 \).

Step 1: Set Equation to Zero

Start by setting the original equation equal to zero: \[ y^{3/2} - 5y = 0 \] This will help in factoring the terms to find solutions for \( y \).

Step 2: Factor Out the Common Term

Factor out the common term \( y \) from the expression:\[ y (y^{1/2} - 5) = 0 \] This tells us that either \( y = 0 \) or \( y^{1/2} - 5 = 0 \).

Step 3: Solve for y

Approach each factor separately: 1. For the factor \( y = 0 \), the solution is simply \( y = 0 \).2. For \( y^{1/2} - 5 = 0 \), solve for \( y \): \( y^{1/2} = 5 \) Square both sides to solve for \( y \): \[ y = (5)^2 = 25 \]

Step 4: Validate Solutions

Make sure both solutions satisfy the original equation:1. For \( y = 0 \): \( 0^{3/2} = 5 \, \times \, 0 \) holds true as 0 on both sides.2. For \( y = 25 \): Compute \( 25^{3/2} = 125 \) and \( 5 \, \times \, 25 = 125 \), thus both sides are equal. Thus, both solutions are valid.

Key Concepts

Equations and Their Role in Problem Solving

Equations form the backbone of algebra and help us establish relationships between different mathematical expressions. An equation is essentially a statement that shows the equality of two expressions. In the problem you're dealing with, the equation is originally given as \( y^{3 / 2} = 5y \).

To find the solution, it's crucial to manipulate this equation into a form that can be solved easily. The goal is to identify the values of the variable \( y \) that make the equation true. To make the equation more manageable, we start by setting it to zero:

• Subtract \( 5y \) from both sides: \( y^{3/2} - 5y = 0 \).

This transformation is vital for further simplification and to apply techniques like factoring.

Factoring as a Key Problem-Solving Technique

Factoring is a valuable technique used to simplify expressions and solve equations. Once you have the equation \( y^{3/2} - 5y = 0 \), the next logical step is to apply factoring. This involves breaking down the expression into simpler components that, when multiplied together, give the original equation.

In this example, you notice a common factor of \( y \):

• Factor it out from each term: \( y(y^{1/2} - 5) = 0 \).

Factoring allows the equation to be broken into two parts: \( y = 0 \) and \( y^{1/2} - 5 = 0 \). These factored components make it easier to find solutions, as you essentially have two separate equations to solve.

Solution Validation for Accurate Results

After finding the potential solutions through the factoring process, it is important to check if they truly satisfy the original equation. Validation ensures that no errors have occurred during solving, particularly when powers and radicals are involved. For the equation such as \( y^{3/2} = 5y \), you've found two solutions: \( y = 0 \) and \( y = 25 \).

Let's validate these solutions:

• When \( y = 0 \), substitute back into the original equation: \( 0^{3/2} = 5 \, \times \, 0 \). Both sides equal 0, so \( y = 0 \) is valid.
• When \( y = 25 \), substitute: \( 25^{3/2} = 125 \) and \( 5 \, \times \, 25 = 125 \). Again, both sides are equal, validating \( y = 25 \).

Through validation, you confirm that the solutions are indeed correct and applicable to the problem.

Understanding Radicals in Equations

Radicals can sometimes make solving equations appear challenging, but understanding their properties makes the process much easier. A radical involves a root of a number, commonly a square root or cube root. In the context of the given problem, \( y^{3/2} \) is a radical expression where you are taking the square root of \( y \) raised to the third power.

During the solving process, simplifying radical expressions is critical:

• To isolate \( y \) in \( y^{1/2} - 5 = 0 \), first recognize \( y^{1/2} \) as the square root of \( y \).
• Then, solve \( y^{1/2} = 5 \) by squaring both sides: \( y = 25 \).

The clear understanding of how to manipulate radicals helps find accurate solutions and apply necessary algebraic transformations to simplify equations. By breaking down the radical expressions clearly, you simplify complex algebraic tasks into more straightforward steps.