Problem 27 - College Algebra with Corequisite Support

Question

For the following exercises, solve the quadratic equation by completing the square. Show each step. $$ x^{2}-6 x=13 $$

Step-by-Step Solution

Answer

The solutions are \(x = 3 + \sqrt{22}\) and \(x = 3 - \sqrt{22}\).

Step 1: Move the Constant

First, move the constant term from the left side of the equation to the right side. We have the equation:\[ x^2 - 6x = 13 \] This remains the same since there is no constant term on the left at this point.

Step 2: Find the Completing Square Term

Find the term needed to complete the square. Take half of the coefficient of \(x\), which is \(-6\), divide it by 2 to get \(-3\), and then square it:\[ \left(\frac{-6}{2}\right)^2 = 9 \] Add this square to both sides of the equation to balance it.

Step 3: Add the Square Term to Both Sides

Add 9 to both sides of the equation, resulting in:\[ x^2 - 6x + 9 = 13 + 9 \] Which simplifies to:\[ x^2 - 6x + 9 = 22 \]

Step 4: Rewrite as a Perfect Square

The left-hand side of the equation is now a perfect square trinomial. Rewrite it as:\[ (x - 3)^2 = 22 \]

Step 5: Take the Square Root of Both Sides

Take the square root of both sides to solve for \(x\):\[ x - 3 = \pm \sqrt{22} \]

Step 6: Solve for x

Finally, solve for \(x\) by adding 3 to both sides:\[ x = 3 \pm \sqrt{22} \]

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra and appears in the form \( ax^2 + bx + c = 0 \). In the equation \( x^2 - 6x = 13 \), the standard form is obtained by rearranging it as \( x^2 - 6x - 13 = 0 \). The highest degree term in this equation is the \( x^2 \), which means it's a quadratic term. Understanding the basic structure of quadratic equations is crucial because they can be solved using various methods, such as factoring, completing the square, or using the quadratic formula. By identifying the coefficients, you can then apply the most suitable method for solving, as we will see further.

Solving Equations

Solving equations refers to the process of finding the value(s) of the variable(s) that satisfy the equation. For quadratic equations, it means finding the values of \( x \) that make the equation true. In this exercise, we solve \( x^2 - 6x = 13 \) by completing the square. This initially involves rearranging the equation to isolate terms involving \( x \) on one side. You'll see these steps more clearly as we go over completing the square. Once you learn solving quadratic equations this way, you'll gain a powerful tool to handle them even when factoring is not feasible.

Perfect Square Trinomial

A perfect square trinomial is a special type of quadratic expression that comes from squaring a binomial, such as \((x - 3)^2\). A trinomial is a polynomial with three terms. When completing the square for \( x^2 - 6x + 9 \), this trinomial is simplified to \((x - 3)^2\). To form a perfect square trinomial, you add value to make the left side of the equation equal to a binomial squared. In this exercise, we added \( 9 \) to \( x^2 - 6x \) because \( 9 \) is \( (-3)^2 \), which converts the left side into \((x - 3)^2\). This transformation allows us to directly apply the square root method later.

Square Root Method

The square root method solves an equation by taking the square root of both sides, which is applicable when one side of the equation is a perfect square. After we converted \( x^2 - 6x + 9 \) into a perfect square trinomial, \((x - 3)^2 = 22\), we apply this method. Solve it by taking the square root, resulting in two possibilities for \( x \) due to the \( \pm \) operations: \( x - 3 = \pm \sqrt{22} \). Adding 3 to both sides then gives us \( x = 3 \pm \sqrt{22} \). This method is useful whenever you have perfect squares, providing a straightforward means to find solutions.