Problem 65 - A Graphical Approach to College Algebra

Question

Solve each quadratic equation by completing the square. $$x(x-1)=3$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{2} \pm \frac{\sqrt{13}}{2}\).

Step 1: Expand the Equation

First, expand the left side of the equation to achieve a standard quadratic form. The given equation is:\[ x(x - 1) = 3 \]Expanding it gives:\[ x^2 - x = 3 \]

Step 2: Rearrange the Equation

Move all terms to one side to set the equation to zero and prepare for completing the square.\[ x^2 - x - 3 = 0 \]

Step 3: Completing the Square

To complete the square, focus on the quadratic and linear terms, \(x^2 - x\). Take half of the coefficient of \(x\), square it, and add this term to both sides.1. Take half of \(-1\) (coefficient of \(x\)): \(-\frac{1}{2}\).2. Square it: \( \left(-\frac{1}{2}\right)^2 = \frac{1}{4} \).Add \(\frac{1}{4}\) to both sides:\[ x^2 - x + \frac{1}{4} = 3 + \frac{1}{4} \]Rewrites as:\[ (x - \frac{1}{2})^2 = \frac{13}{4} \]

Step 4: Solve for x

Now, find the square root of both sides of the equation:\[ \sqrt{(x - \frac{1}{2})^2} = \pm \sqrt{\frac{13}{4}} \]Which simplifies to:\[ x - \frac{1}{2} = \pm \frac{\sqrt{13}}{2} \]Finally, solve for \(x\) by adding \(\frac{1}{2}\) to both sides:\[ x = \frac{1}{2} \pm \frac{\sqrt{13}}{2} \]Separate the solutions:\[ x = \frac{1}{2} + \frac{\sqrt{13}}{2} \] and \[ x = \frac{1}{2} - \frac{\sqrt{13}}{2} \]

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in algebra and appear in various mathematical contexts. They are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. These equations are called "quadratic" because the highest degree of the variable \( x \) is 2. This implies there is a squared term in the equation.

Understanding quadratic equations is crucial, as they often appear in problems involving projectile motion, area calculations, and more. Despite their complexity, there are systematic approaches to solving them, such as factoring, using the quadratic formula, and completing the square. Each method has its advantages depending on the equation's structure.

Standard Quadratic Form

Before solving a quadratic equation, it's often helpful to write it in the standard quadratic form. This form is \( ax^2 + bx + c = 0 \), making it straightforward to apply various solution strategies. Converting an equation to this form involves collecting all terms on one side and simplifying the expression.

Consider the equation originally given as \( x(x-1) = 3 \). To bring it to standard form:

• First, expand the left side to get \( x^2 - x \).
• Subtract 3 from both sides to move all terms to one side, giving \( x^2 - x - 3 = 0 \).

Now, the equation is in standard quadratic form, ready for further solving methods like completing the square. This preparation is essential for handling the equation systematically and identifying the best solution path.

Solving Quadratics

One of the ways to solve quadratic equations is by completing the square, which is particularly useful when factoring is complex or when the quadratic formula is not preferred.

Completing the square involves creating a perfect square trinomial from the quadratic and linear terms. Here's how to do it:

• Take the equation in standard form, \( x^2 - x - 3 = 0 \).
• Focus on \( x^2 - x \), the quadratic and linear terms.
• Take half of the coefficient of the linear term, \( -1 \), which is \( -\frac{1}{2} \), and square it to get \( \frac{1}{4} \).
• Add \( \frac{1}{4} \) to both sides of the equation: \( x^2 - x + \frac{1}{4} = 3 + \frac{1}{4} \).
• This results in \( (x - \frac{1}{2})^2 = \frac{13}{4} \).

Once in this form, solve for \( x \) by taking the square root of both sides and isolating \( x \), leading to the solutions:

• \( x = \frac{1}{2} + \frac{\sqrt{13}}{2} \).
• \( x = \frac{1}{2} - \frac{\sqrt{13}}{2} \).

Completing the square is a powerful technique in algebra because it not only provides solutions to quadratic equations but also helps in understanding their geometrical interpretations.