Problem 23 - Algebra and Trigonometry

Question

Solve the equation by completing the square. \(4 x^{2}-x=0\)

Step-by-Step Solution

Answer

Solutions are \(x = 0\) and \(x = \frac{1}{4}\).

Step 1: Move the constant term

The given equation is \(4x^2 - x = 0\). Our first step is to try to set up the equation such that the left side can form a perfect square. In this equation, there is no constant term, so we don't need to move it to the other side. So, the equation remains \(4x^2 - x = 0\).

Step 2: Factor out the leading coefficient

To make completing the square easier, we will factor out the 4 from the left side of the equation, since 4 is the leading coefficient of \(x^2\). The equation \(4x^2 - x = 0\) becomes \(4(x^2 - \frac{1}{4}x) = 0\), after factoring 4 from the left side.

Step 3: Rearrange to complete the square

To complete the square inside the parentheses, we'll need the middle term to fit the pattern \((x - a)^2 = x^2 - 2ax + a^2\). Take half of the coefficient of \(x\) which is \(\frac{1}{4}\), square it to get \(\left(\frac{1}{8}\right)^2 = \frac{1}{64}\), and add and subtract it within the parentheses: \(4(x^2 - \frac{1}{4}x + \frac{1}{64} - \frac{1}{64}) = 0\).

Step 4: Simplify and solve the equation

Substitute back and simplify the expression: \(4((x - \frac{1}{8})^2 - \frac{1}{64}) = 0\). Distribute the 4: \(4(x - \frac{1}{8})^2 - \frac{4}{64} = 0\). Simplify \(\frac{4}{64}\) to \(\frac{1}{16}\): \(4(x - \frac{1}{8})^2 - \frac{1}{16} = 0\).

Step 5: Solve for \(x\)

Add \(\frac{1}{16}\) to both sides: \(4(x - \frac{1}{8})^2 = \frac{1}{16}\). Divide by 4: \((x - \frac{1}{8})^2 = \frac{1}{64}\). Take the square root of both sides: \(x - \frac{1}{8} = \pm\frac{1}{8}\). Solve for \(x\): \(x = \frac{1}{8} + \frac{1}{8}\) or \(x = \frac{1}{8} - \frac{1}{8}\). Thus, \(x = \frac{1}{4}\) or \(x = 0\).

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\). They represent a parabola when graphed. Quadratic equations can appear in many different formats, but the essence is the same: a quadratic expression equals zero. This type of equation features an \(x^2\) term, which makes it unique from linear equations. In the given problem, the quadratic equation is structured as \(4x^2 - x = 0\). Though it doesn't have a constant \(c\) term, it still qualifies as a quadratic because the highest power of \(x\) is 2.

Solving quadratic equations can be done using various methods, including factoring, completing the square, and using the quadratic formula. Each of these methods has its scenarios where they are more efficient or insightful. Let's explore one particular method: completing the square. It is especially useful when the equation cannot be easily factored.

Factoring

Factoring is a method to solve quadratic equations by expressing the quadratic in the form of \((x - p)(x - q) = 0\). With this form, the solutions to the equation are simply the values of \(x\) that make each factor zero. The given expression \(4x^2 - x = 0\) can be viewed in factored form by pulling out the greatest common factor.

In our example, the greatest common factor is \(x\). So, the equation becomes \(x(4x - 1) = 0\). This clearly shows the two potential solutions: \(x = 0\) or \(4x - 1 = 0\). From solving \(4x - 1 = 0\), we get \(x = \frac{1}{4}\).

This demonstrates factoring as a powerful tool to simplify and find solutions for quadratic equations. However, certain quadratics may not be easily factorable, which is where completing the square might come into play.

Perfect Square Trinomial

A perfect square trinomial is a special quadratic form \((x - a)^2 = x^2 - 2ax + a^2\). Recognizing this pattern helps in transforming quadratics to make them easier to solve. The trick is to manipulate the quadratic to complete "the square," thus turning it into a perfect square trinomial.

To complete the square for \(x^2 - \frac{1}{4}x\), we need to find a value that completes the pattern of a perfect square. Half the coefficient of \(x\) (which is \(-\frac{1}{4}\)), is \(-\frac{1}{8}\). Squaring this gives \(\frac{1}{64}\), enabling us to write \((x - \frac{1}{8})^2\). By adding and subtracting \(\frac{1}{64}\), we ensure the expression inside the parentheses forms a perfect square trinomial, crucial for solving by completing the square.

Solving Equations

Solving equations, particularly quadratics, revolves around finding values for \(x\) that satisfy the equation. When a quadratic is not easily factorable, completing the square offers a systematic approach. Using this method, a quadratic equation can often be transformed into an easily solvable square.

For our equation \((x - \frac{1}{8})^2 = \frac{1}{64}\), we solved it by taking the square root of both sides. This yields \(x - \frac{1}{8} = \pm \frac{1}{8}\). By solving these two linear equations, \((x - \frac{1}{8} = \frac{1}{8}\) and \(x - \frac{1}{8} = -\frac{1}{8})\), we found the solutions \(x = \frac{1}{4}\) and \(x = 0\).

• Isolating the quadratic term to complete the square simplifies a solution process.
• Solving the resulting perfect square leads directly to real solutions.

By making these transformations and solving them step-by-step, even complex quadratics become manageable.