Problem 10 - A Graphical Approach to College Algebra

Question

Answer each question. Only one of the following equations does not require Step 1 of the method for completing the square. Which one is it? Solve it. A. \(3 x^{2}-17 x-6=0\) B. \((2 x+5)^{2}=7\) C. \(x^{2}+x=12\) D. \((3 x+1)(x-7)=0\)

Step-by-Step Solution

Answer

Equation D, solved as \(x=-\frac{1}{3}\) and \(x=7\), does not require completing the square.

Step 1: Identify Equation That Doesn't Need Completing the Square

To determine which equation does not require completing the square, look for equations already set up to use a different method. Option D, \((3x+1)(x-7)=0\), is a product of two binomials set to zero, which can be solved using the Zero Product Property.

Step 2: Use Zero Product Property

For the equation \((3x+1)(x-7)=0\), set each factor equal to zero: \(3x+1=0\) and \(x-7=0\). This gives two simple linear equations to solve.

Step 3: Solve the First Linear Equation

For \(3x+1=0\), subtract 1 from both sides to get \(3x=-1\). Then, divide both sides by 3 to solve for \(x\), resulting in \(x=-\frac{1}{3}\).

Step 4: Solve the Second Linear Equation

For \(x-7=0\), add 7 to both sides to solve for \(x\). This yields \(x=7\).

Step 5: Combine Solutions

The solutions to the equation \((3x+1)(x-7)=0\) are \(x=-\frac{1}{3}\) and \(x=7\). Both solutions satisfy the initial equation.

Key Concepts

Completing the Square

Completing the square is a method used in algebra to solve quadratic equations by converting them into a perfect square trinomial. This technique makes the equation easier to solve by allowing us to isolate the variable more simply. To complete the square, follow these steps:

• Start with a quadratic equation in the form of \( ax^2 + bx + c = 0 \).
• Move the constant term to the other side of the equation: \( ax^2 + bx = -c \).
• If \( a eq 1 \), divide every term by \( a \) to simplify to \( x^2 + \frac{b}{a} x = \frac{-c}{a} \). This step is crucial for easy manipulation.
• Find the value that completes the square: take \( \frac{b}{2a} \), square it, and add this value to both sides.
• This transforms the left side into a perfect square trinomial, expressed as \((x + p)^2\), and simplifies your equation.

The beauty of completing the square is it directly leads to solving for \( x \) when using the formula for a perfect square. It gives insight into the geometry of quadratic equations, specifically their graph's vertex.

Quadratic Equations

Quadratic equations are polynomials with a degree of 2, which means their highest exponent is squared. They generally have the standard form \( ax^2 + bx + c = 0 \). Quadratic equations can have two real solutions, one real solution, or two complex solutions, depending on the discriminant \( b^2 - 4ac \). Key methods to solve these equations include:

• Factoring: If the quadratic can be expressed as a product of two binomials, set each binomial to zero and solve separately as done in Option D of the original exercise.
• Using the Quadratic Formula: This provides solutions for any quadratic equation using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). It is a reliable method for cases where factoring is complicated.
• Completing the Square: As previously discussed, this method reshapes the equation into a solvable form by creating a perfect square trinomial.

Understanding how to work with quadratic equations is fundamental for solving a broad range of problems in algebra, from simple maths problems to more complex physics problems.

Binomial Equations

Binomials are algebraic expressions containing two terms, made from the sum or difference of monomials. When these two terms are multiplied to zero, as seen in the equation \((3x+1)(x-7)=0\), they form a binomial equation.Solving binomial equations often involves the use of the Zero Product Property, which states if \( ab = 0 \), then either \( a = 0 \) or \( b = 0 \), or both. In the given problem:

• Each binomial factor is set to zero, creating two separate linear equations: \( 3x + 1 = 0 \) and \( x - 7 = 0 \).
• Solve each equation independently to find possible values for \( x \) that satisfy the original equation. Here, \( x = -\frac{1}{3} \) and \( x = 7 \).

This approach is straightforward and is one of the simplest methods for solving polynomial equations by breaking them down into simpler parts. Binomial equations and properties like these provide foundational skills useful for both basic algebra and more complex mathematical concepts.