Problem 79 - Algebra 2

Question

Solve each equation by factoring. \(18 u^{2}-3 u=1\)

Step-by-Step Solution

Answer

The solutions are \( u = -\frac{1}{6} \) and \( u = \frac{1}{3} \).

Step 1: Move All Terms to One Side

Start by moving all terms to one side of the equation to set it equal to zero. Subtract 1 from both sides:\[ 18u^2 - 3u - 1 = 0 \]

Step 2: Recognize the Quadratic Form

We have a quadratic equation in the form \( au^2 + bu + c \), where \( a = 18 \), \( b = -3 \), and \( c = -1 \). We will factor it next.

Step 3: Use the AC Method

To factor the quadratic, employ the AC method. Multiply \( a \) and \( c \): \( 18 \times -1 = -18 \). We need two numbers that multiply to -18 and add up to \( -3 \). These numbers are \( -6 \) and \( 3 \).

Step 4: Rewrite and Group Terms

Rewrite the middle term, \( -3u \), using the numbers found: \[ 18u^2 - 6u + 3u - 1 = 0 \] Group the terms:\[ (18u^2 - 6u) + (3u - 1) = 0 \]

Step 5: Factor by Grouping

Factor out the greatest common factor (GCF) from each group:\[ 6u(3u - 1) + 1(3u - 1) = 0 \] Factor out the common binomial factor \( (3u - 1) \): \[ (6u + 1)(3u - 1) = 0 \]

Step 6: Solve for u

Set each factor equal to zero and solve for \( u \):1. \( 6u + 1 = 0 \) - Subtract 1: \( 6u = -1 \) - Divide by 6: \( u = -\frac{1}{6} \) 2. \( 3u - 1 = 0 \) - Add 1: \( 3u = 1 \) - Divide by 3: \( u = \frac{1}{3} \)

Key Concepts

Quadratic Equation

A quadratic equation is an algebraic expression set equal to zero, featuring variables raised to the second power. It comes in the standard form of \[ ax^2 + bx + c = 0 \]where:

• \(a\), \(b\), and \(c\) are constants
• \(x\) is the unknown variable
• \(a eq 0\) to ensure it's truly quadratic

Quadratic equations appear in various real-world situations such as physics, engineering, and finance. Understanding their structure is crucial for solving and applying them effectively.

AC Method

The AC method is a popular technique used to factor quadratic equations when direct factoring isn't straightforward. This method involves:

• Multiplying the coefficient of the quadratic term (\(a\)) by the constant term (\(c\)).
• Finding two numbers that multiply to the result (\(ac\)) but also add up to the middle term's coefficient \(b\).

This approach transforms the process into finding pairs of numbers rather than arbitrarily guessing. It significantly eases the task of cracking tougher equations for students.

Factoring by Grouping

Factoring by grouping is a strategic method to break down polynomial expressions into factors. After identifying the required pair of numbers using the AC method, rewrite the middle term and group terms as follows:

• Take terms from the new equation and group them into pairs.
• For each group, extract the greatest common factor (GCF).
• Detect identical factors across groups to enable further simplification.

This systematic approach turns a complex problem into smaller, more manageable pieces, making it easier to solve.

Solving Equations

Solving equations like quadratics involves finding the values of the unknown variable that satisfy the equation. After factoring, the striving goal is to set each factor equal to zero and solve as follows:

• Split the equation: Ensure each factor can be equated to zero independently.
• Isolate the variable by performing basic algebraic operations like addition or subtraction.
• Ensure you divide appropriately to solve for the unknown variable.

This process lets you pinpoint specific solutions within the equation's framework, an essential skill in mathematics.