Problem 18 - Algebra 2 and Trigonometry

Question

In \(18-35,\) find each common solution algebraically. Express irrational roots in simplest radical form. $$ \begin{array}{l}{y=x^{2}-2 x} \\ {y=3 x}\end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Set the Equations Equal

To find the common solutions of the two equations, set the two expressions for \(y\) equal to each other: \[ x^2 - 2x = 3x \]

Step 2: Simplify the Equation

Move all terms to one side of the equation to form a quadratic equation: \[ x^2 - 2x - 3x = 0 \]Simplify it to: \[ x^2 - 5x = 0 \]

Step 3: Factor the Equation

Factor out an \(x\) from the quadratic equation: \[ x(x - 5) = 0 \]

Step 4: Find the Solutions

Set each factor equal to zero and solve for \(x\): 1. \(x = 0\) 2. \(x - 5 = 0 \implies x = 5\)

Step 5: Verify by Substitution

Substitute \( x = 0 \) and \( x = 5 \) back into the original equations to verify if both values satisfy both equations: For \(x = 0:\) - \(y = 0^2 - 2 \times 0 = 0\) - \(y = 3 \times 0 = 0\) For \(x = 5:\) - \(y = 5^2 - 2 \times 5 = 25 - 10 = 15\) - \(y = 3 \times 5 = 15\)Both \(x = 0\) and \(x = 5\) satisfy the original equations.

Key Concepts

Factoring

Factoring is a crucial step in solving many algebraic equations, especially quadratic equations. It simplifies finding solutions by breaking down complicated expressions into simpler components that can be easily solved. In the exercise given, the equation was transformed to a standard quadratic form by first setting the two expressions for \(y\) equal to each other, resulting in \( x^2 - 2x = 3x \). By bringing all terms to one side, we get \( x^2 - 5x = 0 \).

The next step involves factoring the equation. Notice that both terms in the equation have an \(x\) in common. Thus, we factor out \(x\), leading to \( x(x - 5) = 0 \). This factorization is essential because now the solutions can be easily determined by setting each factor equal to zero. Factoring reduces the complexity by breaking down the equation into parts that are more straightforward to solve.

Solving Equations

Solving equations is the process of finding the values of the variable that make the equation true. After factoring the quadratic equation \( x(x - 5) = 0 \), we easily find the solutions by setting each factor to zero.

• First, solve \(x = 0\), a factor easily visible from the equation as one possible solution.
• Next, solve \(x - 5 = 0\), which gives the solution \(x = 5\).

These values, \(x = 0\) and \(x = 5\), satisfy the equation \( x^2 - 5x = 0 \) because substituting either of these values into the equation results in a true statement. Always check your solutions by substituting them back into the original equations to ensure they fit both forms of \(y\). In this exercise, both solutions work and verify correctly by substitution back into the original equations, confirming their validity.

Algebraic Solutions

Algebraic solutions are essential when working with equations, as they provide the concrete values for variables that satisfy the given conditions. Solving algebraically involves logical steps such as factoring, simplifying, and substituting back to verify solutions. When solving the quadratic equation derived from\( y = x^2 - 2x \) and \( y = 3x \), initial steps include equating, rearranging, and factoring.

Once allotted into manageable expressions, such as \( x(x - 5) = 0 \), the algebraic solutions \(x = 0\) and \(x = 5\) emerge clearly. It's important to remember the final step, which involves verifying your solutions by substituting back into the original equations. This reassures the solver that both \(x\) values fit and uphold the conditions set by both equations. Practicing these algebraic techniques enhances problem-solving skills and prepares students for more complex algebra challenges.