Problem 44 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ x^{2}=5 x $$

Step-by-Step Solution

Answer

The solutions are \(x = 0\) and \(x = 5\).

Step 1: Set the Equation to Zero

Start by rewriting the equation in the form where one side is zero. This helps in factoring the equation later. The given equation is \(x^2 = 5x\). Subtract \(5x\) from both sides to get: \(x^2 - 5x = 0\).

Step 2: Factor the Equation

Factor the equation \(x^2 - 5x = 0\). Look for common factors in both terms. Both terms share \(x\), so factor out \(x\): \(x(x - 5) = 0\).

Step 3: Solve Each Factor Equal to Zero

Set each factor equal to zero and solve for \(x\). The equation \(x(x-5) = 0\) gives us two factors: \(x = 0\) and \(x - 5 = 0\).1. For \(x = 0\), the solution is simply \(x = 0\).2. For \(x - 5 = 0\), add 5 to both sides to solve: \(x = 5\).

Step 4: Verify the Solutions

Substitute each solution back into the original equation to ensure they solve the equation.1. For \(x = 0\), substitute into \(x^2 = 5x\): \(0^2 = 5(0)\) gives \(0 = 0\), which is true.2. For \(x = 5\), substitute: \(5^2 = 5(5)\) gives \(25 = 25\), which is also true.

Key Concepts

Factoring

Factoring is the process used to break down an algebraic expression into simpler "factors" or parts that, when multiplied, give back the original expression. It is one of the fundamental tools in solving quadratic equations. Here, we applied factoring to the quadratic equation derived from the exercise:
- Original Equation: \(x^2 = 5x\) - Transformed to: \(x^2 - 5x = 0\)
To factor this expression, we look for common factors in each term. In this case, both terms \(x^2\) and \(-5x\) contain \(x\). By factoring \(x\) out, the quadratic equation becomes:

• \(x(x - 5) = 0\)

This means that the equation is broken down into two parts: \(x\) and \(x - 5\). Factoring helps identify the individual terms that, when equated to zero, give us the solutions to the equation.

Zero Product Property

The Zero Product Property is a crucial principle in algebra that states: if the product of any two numbers equals zero, then at least one of the factors must be zero. This property provides a straightforward method to solve factorable equations like the one we have here.After factoring the equation \(x(x - 5) = 0\), we apply the Zero Product Property. This results in two possible equations:

• \(x = 0\)
• \(x - 5 = 0\)

By solving each equation individually, we find the potential solutions to the original equation:- For \(x = 0\), the equation already indicates the solution.- For \(x - 5 = 0\), solving gives us \(x = 5\).
The Zero Product Property thus simplifies the task of dealing with quadratic equations by breaking them down into solvable parts.

Algebraic Equations

Algebraic equations are mathematical statements where two expressions are set equal to each other. Solving these equations requires finding the value(s) of the variable(s) that satisfy the equality. Let's see how this applies to the given example.The initial equation from the problem is a quadratic equation: - \(x^2 = 5x\)
Our goal is to solve for \(x\).

• First, we set the equation to zero: \(x^2 - 5x = 0\).
• Then, we factor the equation to separate it into simpler terms: \(x(x - 5) = 0\).

Using the solutions from the Zero Product Property, we determine that \(x = 0\) or \(x = 5\) are both solutions. Substituting these values back into the original equation confirms their correctness:- Substituting \(x = 0\) results in \(0 = 0\).- Substituting \(x = 5\) results in \(25 = 25\).Thus, these solutions are valid, reaffirming our algebraic manipulations in reaching them.