Problem 67 - A Graphical Approach to College Algebra

Question

Solve each quadratic equation by completing the square. $$2 x^{2}-x+3=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{4} \pm \frac{\sqrt{23}}{4}i\).

Step 1: Move Constant to the Other Side

Start by moving the constant term (3) to the right side of the equation to focus on the terms involving x. Subtract 3 from both sides: \[2x^2 - x = -3\]

Step 2: Simplify Coefficient of Quadratic Term

Divide the whole equation by 2 so that the coefficient of \(x^2\) becomes 1. This will make completing the square easier:\[x^2 - \frac{1}{2}x = -\frac{3}{2}\]

Step 3: Complete the Square

To complete the square, take half of the coefficient of \(x\), square it, and add it to both sides. Here, the coefficient of \(x\) is \(-\frac{1}{2}\). Half of that is \(-\frac{1}{4}\), and after squaring, you get:\[\left(-\frac{1}{4}\right)^2 = \frac{1}{16}\]Add \(\frac{1}{16}\) to both sides:\[x^2 - \frac{1}{2}x + \frac{1}{16} = -\frac{3}{2} + \frac{1}{16}\]

Step 4: Simplify and Factor

The left side is a perfect square trinomial, and can be written as a square of a binomial:\[\left(x - \frac{1}{4}\right)^2 = -\frac{24}{16} + \frac{1}{16} = -\frac{23}{16}\]

Step 5: Solve for x

Take the square root of both sides to solve for \(x\):\[x - \frac{1}{4} = \pm \sqrt{-\frac{23}{16}}\]Since \(\sqrt{-1} = i\) (the imaginary unit), we have:\[x - \frac{1}{4} = \pm \frac{\sqrt{23}}{4}i\]

Step 6: Isolate x

Add \(\frac{1}{4}\) to both sides to solve for \(x\):\[x = \frac{1}{4} \pm \frac{\sqrt{23}}{4}i\]

Key Concepts

Quadratic Equations

Quadratic equations are expressions of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown variable. These equations are fundamental in algebra and appear often in various forms in mathematical problems. The goal when solving a quadratic equation is to find the values of \(x\) that make the equation true. There are different methods for solving them such as factoring, using the quadratic formula, or completing the square.

The general method of completing the square involves rearranging the equation to form a perfect square trinomial on one side of the equation. This method is very useful because it directly leads us to the vertex form of a parabola, opening ways to better understand its graphical representation. It also links nicely with other algebraic structures and methods.

The key steps include:

• Moving the constant term to the opposite side
• Dividing through by the coefficient of \(x^2\) if it's not 1
• Completing the square by adding the necessary term to both sides
• Performing algebraic manipulations to solve for \(x\)

Imaginary Numbers

Imaginary numbers are numbers that result from taking the square root of a negative number. The symbol "\(i\)" is used to represent the square root of -1, known as the imaginary unit. This concept extends the real number system and is foundational for complex numbers, which are numbers of the form \(a + bi\), where \(a\) and \(b\) are real numbers.

In solving the quadratic equation derived from completing the square, we encounter \(\pm \sqrt{-\frac{23}{16}}\). Here, since we are taking a square root of a negative number, imaginary numbers are involved:

• \(\sqrt{-1} = i\)
• The expression then becomes \(\pm \frac{\sqrt{23}}{4}i\)

This expands our ability to solve equations and model more complex phenomena which can't be explained with just real numbers.

Perfect Square Trinomial

A perfect square trinomial is a specific type of quadratic expression that can be expressed in the form \((a+b)^2 = a^2 + 2ab + b^2\). This special structure makes it easy to factor the trinomial into a binomial raised to the second power.

In the context of completing the square, we purposefully structure one side of the equation as a perfect square trinomial to make the equation more easily solvable. For instance, consider the step where we obtained \(x^2 - \frac{1}{2}x + \frac{1}{16}\) from our original expression. This becomes \((x - \frac{1}{4})^2\), a format that readily allows further simplifications.

• This step simplifies the equation to a form that makes extracting \(x\) more straightforward after taking square roots.
• The method assists in deriving a clearer, condensed format while solving quadratic equations.

This concept of transforming terms into perfect squares is a foundational algebraic technique, particularly useful for simplifying and solving equations.