Problem 27 - College Algebra

Question

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 Constant to Right Side

Start by ensuring the quadratic equation is in the form of \( ax^2 + bx = c \). In this case, it is \( x^2 - 6x = 13 \), so no need to rearrange.

Step 2: Create a Perfect Square on the Left

To complete the square, take half of the coefficient of \( x \) (which is -6), and square it. Half of -6 is -3, and \((-3)^2 = 9\). Add 9 to both sides of the equation to maintain equality: \( x^2 - 6x + 9 = 13 + 9 \).

Step 3: Simplify the Equation

Simplify both sides: \( x^2 - 6x + 9 = 22 \) becomes \( (x-3)^2 = 22 \) because \( x^2 - 6x + 9 \) is a perfect square trinomial.

Step 4: Solve for x

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

Step 5: Final Step, Isolate 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 type of polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants and \( a eq 0 \). The highest power of the variable, usually \( x \), is 2. Quadratic equations can appear in various forms and require specific techniques to solve, such as factoring, using the quadratic formula, or completing the square.

In our specific exercise, we have the quadratic equation \( x^2 - 6x = 13 \). Note that it can be rewritten by moving all terms to one side to match the canonical zero-based form, allowing easier application of solving methods.

Perfect Square Trinomial

A perfect square trinomial is a special quadratic expression that can be expressed as the square of a binomial. For example, \( (x-3)^2 \) expands to \( x^2 - 6x + 9 \). Noticing these patterns is a key to understanding the completing the square method.

To transform \( x^2 - 6x \) into a perfect square trinomial, identify the coefficient of the \( x \)-term. Divide it by two and square the result. Since \(-6/2 = -3\) and \((-3)^2 = 9\), adding 9 creates \( x^2 - 6x + 9 \). This process makes the left side a perfect square trinomial, simplifying our problem.

Algebraic Manipulation

Algebraic manipulation involves reworking equations to make them easier to solve. Techniques include adding, subtracting, multiplying, or dividing both sides of the equation by the same non-zero number. By using these techniques strategically, we can preserve equality while reshaping the equation.

In our example, by adding 9 to both sides, we maintain the equation's equality while transforming its structure for easier solving:

• Original: \( x^2 - 6x = 13 \)
• Addition: \( x^2 - 6x + 9 = 22 \)
• Factoring: \((x-3)^2 = 22\)

Each step uses algebraic manipulation to make the equation more tractable for further solution steps.

Equation Solving Steps

Solving quadratic equations, especially by completing the square, follows a systematic approach. This ensures that each step logically progresses toward the solution. Here's a concise breakdown of the steps taken in the exercise:

• Start by arranging the equation in a standard form if needed. For our example, it's already in the right form.
• Create a perfect square trinomial on one side by adding the square of half the \( x \)-coefficient to both sides. In our problem, this leads to \( x^2 - 6x + 9 = 22 \).
• Factor the trinomial into a binomial square \((x-3)^2 = 22\).
• Take the square root of both sides to find \( x - 3 = \pm \sqrt{22} \).
• Finally, isolate \( x \) by adding 3 to both sides, yielding \( x = 3 \pm \sqrt{22} \).

Each step builds on the last, cementing the logical path to the solution.