Problem 84 - Precalculus with Limits

Question

In Exercises 77-84, solve for \(x\). \(\left| \begin{array}{c} x+4 & -2 \\ 7 & x-5 \end{array} \right| = 0\)

Step-by-Step Solution

Answer

The solutions for \(x\) are \(3\) and \(-2\)

Step 1: Setting up the Determinant

For the 2x2 matrix, the determinant is defined as the product of the values in the main diagonal subtracted by the product of the values in the other diagonal. So, the first step is to set up this determinant for our matrix. That gives us \( (x+4) \cdot (x-5) - (-2) \cdot 7 \).

Step 2: Simplifying the Expression

We simplify the expression by multiplying the values and combining like terms. This gives us \(x^2 - x - 20 + 14\). If we further simplify this by combining constant terms, it results in \(x^2 - x - 6\).

Step 3: Setting the Equation Equal to Zero

We set this equation equal to zero to solve for \(x\), as per the requirement of the exercise: \(x^2 - x - 6 = 0\).

Step 4: Factoring the Quadratic Equation

We factor the quadratic equation into two binomials: \((x - 3)(x + 2) = 0\).

Step 5: Solving for \(x\)

Finally, we solve for \(x\) by setting each factor equal to zero: \(x - 3 = 0\) or \(x + 2 = 0\). Solving each of these equations for \(x\) gives us \(x = 3\) and \(x = -2\).

Key Concepts

Quadratic Equations

Quadratic equations are an essential topic in algebra. They are polynomial equations of degree two and typically take the form \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). The quadratic equation we encountered in the exercise is \[ x^2 - x - 6 = 0 \]. This particular equation arose from simplifying the determinant of a matrix to solve for \( x \). Quadratic equations can be solved by a variety of methods including factoring, completing the square, or using the quadratic formula. Each method has its own applications and advantages. For instance, factoring is straightforward if the quadratic can be easily expressed as a product of binomials.

Factoring Polynomials

Factoring polynomials is a method used to simplify polynomial expressions by expressing them as a product of simpler polynomials. In the original exercise, we factored the quadratic equation \[ x^2 - x - 6 = 0 \] into two binomials: \[ (x - 3)(x + 2) = 0 \]. To factor successfully, you need to find two numbers that multiply to the constant term (here, \( -6 \)) and add up to the linear coefficient \( -1 \). By observing, we see that \( -3 \) and \( 2 \) meet these conditions:

• \((-3) \times (2) = -6\)
• \((-3) + 2 = -1\)

Factoring is not only a critical skill in algebra but also simplifies many mathematical operations by reducing complex expressions to more manageable terms.

Solving Equations

Solving equations is a fundamental aspect of mathematics. The goal is to find the values of the unknowns that satisfy the equation. In this scenario, we were solving a quadratic equation derived from a matrix determinant. Once the quadratic \[ (x - 3)(x + 2) = 0 \] was factored, the process of solving involved finding the values of \( x \) that make each factor equal to zero:

• Setting \(x - 3 = 0\) results in \(x = 3\).
• Setting \(x + 2 = 0\) gives \(x = -2\).

Each solution satisfies one of the factored equations, meaning both are solutions to the original quadratic equation. Solving equations requires you to manipulate algebraic expressions and ensure that each step logically follows from the previous one, ultimately leading to the solutions that satisfy the original mathematical statement.