Problem 41 - College Algebra

Question

For the following exercises, solve the equation by identifying the quadratic form. Use a substitute variable and find all real solutions by factoring. $$ (x-3)^{2}-4=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = 5\).

Step 1: Rewrite the Equation

Observe the given equation \((x-3)^{2}-4=0\). Notice that it can be written in a form similar to a difference of squares: \((a)^2-(b)^2=0\). Rewrite it as \([((x-3)^2) - (2)^2 = 0]\).

Step 2: Identify the Quadratic Form and Substitute

To simplify, identify a new variable substitution. Let \(y = x - 3\). Then \((x-3)^2\) becomes \(y^2\), and the equation is \(y^2 - 4 = 0\).

Step 3: Express the Equation as a Difference of Squares

Recognize the quadratic form in the equation \(y^2 - 4 = (y - 2)(y + 2) = 0\). This is a difference of squares, where \(a = y\) and \(b = 2\).

Step 4: Solve for the Substitute Variable

Factor the equation \((y-2)(y+2) = 0\). Set each factor equal to zero: \((y-2) = 0\) or \((y+2) = 0\). Solve for \(y\). The solutions are \(y = 2\) and \(y = -2\).

Step 5: Substitute Back to Find Original Variable Solutions

Recall the substitution \(y = x - 3\). Solve for \(x\) for each value of \(y\): \(x - 3 = 2 \Rightarrow x = 5\) and \(x - 3 = -2 \Rightarrow x = 1\).

Step 6: State All Real Solutions

The real solutions to the original equation \((x-3)^2-4=0\) are \(x = 5\) and \(x = 1\).

Key Concepts

Difference of Squares

The concept of the difference of squares is a useful factorization technique that is frequently used in algebra. It involves expressing a quadratic expression of the form \(a^2 - b^2\) as the product of two binomials:

• \( (a - b)(a + b) \)

This is a powerful tool because it allows for the simplification of seemingly complex expressions. In the equation \((x-3)^{2}-4=0\), we can recognize it as fitting the difference of squares pattern. Here, \(a\) is \((x-3)\) and \(b\) is 2 since it can be repositioned into the form \(((x-3)^2) - (2)^2 = 0\). By rewriting this expression using the difference of squares identity, we obtain \((x-3 - 2)(x-3 + 2) = 0\), which simplifies the process of finding solutions.

Variable Substitution

Variable substitution is a technique used to simplify equations and expressions by introducing a new variable to replace a more complex expression. This often makes the equation more manageable and easier to solve.In our exercise, the expression \((x-3)\) is set equal to a new variable, \(y\). Therefore, \((x-3)^2\) becomes \(y^2\). With this substitution, the equation \((x-3)^2 - 4 = 0\) transforms into the simpler form \(y^2 - 4 = 0\). By replacing complex parts of the equation with a variable, we focus on solving the core structure without immediately addressing the complexity of the expression.

Factoring Quadratics

Factoring quadratics is an essential method in algebra to solve quadratic equations. It involves breaking down the equation into simpler expressions that can be solved individually. This method is particularly useful when dealing with polynomial expressions of the form \(ax^2 + bx + c = 0\).In our example, after implementing the difference of squares and variable substitution, the equation was simplified to \(y^2 - 4 = 0\). Recognizing it as a difference of squares, we can factor it as \((y - 2)(y + 2) = 0\). To find the roots of the equation, we solve each factor separately:

• Set \(y - 2 = 0\) which gives \(y = 2\).
• Set \(y + 2 = 0\) which yields \(y = -2\).

These roots correlate back to the original variable through substitution reversal: \(x - 3 = 2\) or \(x - 3 = -2\), leading to the comprehensive solutions \(x = 5\) and \(x = 1\). Factoring thus plays a key role in uncovering the solutions to the original quadratic equation.