Problem 55 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation using any method. Find only real solutions. $$x^{2}-4=0$$

Step-by-Step Solution

Answer

The solutions to the equation \(x^{2}-4=0\) are \(x = 2\) and \(x = -2\).

Step 1: Recognize the Form of the Equation

The provided equation can be identified as a difference of squares. It follows the form of \(a^{2} - b^{2} = (a - b)(a + b)\) where \(a = x\) and \(b = 2\).

Step 2: Factor the Equation

Apply the difference of squares formula to factor the equation. It becomes: \((x - 2)(x + 2) = 0\).

Step 3: Solve the Derived Equations

Having factored the equation, you now have two simple linear equations: \(x - 2 = 0\) and \(x + 2 = 0\). Solving each of these for x gives solutions \(x = 2\) and \(x = -2\) respectively.

Key Concepts

Factoring Quadratics

Factoring quadratics is a critical skill when it comes to solving quadratic equations. A quadratic equation typically takes the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a \) is not zero. The goal of factoring is to break down the original quadratic into a product of simpler binomials. Think of it like breaking a large piece into two smaller chunks that are easier to analyze.

To factor a quadratic, one often looks for two numbers that both add up to \( b \) and multiply to \( ac \). These numbers will be used in the binomials in place of \( b \) when factoring. For instance, if the quadratic were \( x^2 + 5x + 6 \), we'd look for two numbers that add up to 5 and multiply to 6, which are 2 and 3. Therefore, the factored form would be \( (x + 2)(x + 3) = 0 \).

Sometimes quadratics won't factor nicely, in which case alternative methods like completing the square or using the quadratic formula may be used. Remember, factoring is often the fastest and simplest route to finding the solutions for a quadratic equation if the polynomial factors neatly.

Difference of Squares

The difference of squares is a special factoring pattern you'll often encounter when dealing with quadratic equations. It occurs when a quadratic is composed of two terms that are perfect squares themselves, separated by a minus sign. The formula for the difference of squares is \( a^2 - b^2 = (a - b)(a + b) \).

Why is this formula so handy? Because recognizing it allows for quick factoring of an equation that might otherwise seem daunting. For example, in the exercise \( x^2 - 4 = 0 \), the number 4 is a perfect square since \( 2^2 = 4 \). Consequently, the equation fits the pattern of a difference of squares, where \( a = x \) and \( b = 2 \), thus it rapidly factors to \( (x - 2)(x + 2) = 0 \).

This method effectively transforms the solving process into a much simpler one, as it breaks down the equation into linear equations that are more manageable and straightforward to solve. Understanding and identifying the difference of squares is a great strategy for quick and efficient problem-solving in algebra.

Real Solutions of Quadratics

Real solutions of quadratics come into play when we discuss the types of answers we can get from a quadratic equation. Not all quadratics have real solutions; some have complex solutions when the equation involves the square root of a negative number. However, a simple way to predict whether you'll get real solutions is by calculating the determinant (also known as the discriminant) from the coefficients in the quadratic equation's standard form, \( ax^2 + bx + c = 0 \). The determinant is given by \( b^2 - 4ac \).

If the determinant is positive, you'll get two real solutions. If it's zero, you'll have one real solution. However, if the determinant is negative, this indicates that no real solutions exist, and any solutions will be in the form of complex numbers. In the given exercise, there is no need to calculate the determinant as the factoring shows outright that two real solutions exist: \( x = 2 \) and \( x = -2 \). Whenever you can factor the quadratic equation, as in our example, it clearly indicates how many real solutions exist based on the factors derived.