Problem 65 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ \left(x-\frac{1}{8}\right)^{2}=\frac{1}{64} $$

Step-by-Step Solution

Answer

x = \frac{1}{4} or x = 0.

Step 1: Recognize the Square Root Property

To use the square root property, we start with an equation of the form \((x - a)^{2} = b\). Our equation, \(\left(x - \frac{1}{8}\right)^{2} = \frac{1}{64}\), fits this form, where \(a = \frac{1}{8}\) and \(b = \frac{1}{64}\).

Step 2: Apply the Square Root Property

The square root property states that if \(x^2 = k\), then \(x = \pm \sqrt{k}\). Applying this to our equation, \(x - \frac{1}{8} = \pm \sqrt{\frac{1}{64}}\).

Step 3: Simplify the Radical

Simplify the right side \(\sqrt{\frac{1}{64}} = \frac{1}{8}\). Therefore, the equation becomes \(x - \frac{1}{8} = \pm \frac{1}{8}\).

Step 4: Solve for x

We now have two possible equations: \(x - \frac{1}{8} = \frac{1}{8}\) and \(x - \frac{1}{8} = -\frac{1}{8}\). Solve each one separately: \(x = \frac{1}{8} + \frac{1}{8} = \frac{2}{8} = \frac{1}{4}\), and \(x = \frac{1}{8} - \frac{1}{8} = 0\).

Step 5: State the Solutions

The solutions to the original equation are \(x = \frac{1}{4}\) and \(x = 0\).

Key Concepts

Square Root Property

When solving quadratic equations, the square root property is a powerful tool. This property can simplify and solve equations with squared terms. If you have an equation of the form \[ (x - a)^2 = b \], you can apply the square root property to remove the square. This property states that if \[ x^2 = k \], then \[ x = \pm \sqrt{k} \]. Breaking it down, this essentially means we take the square root of both sides, remembering to consider both the positive and negative roots. For example, in the equation \[ (x - \frac{1}{8})^2 = \frac{1}{64} \], we identify \(a = \frac{1}{8}\) and \(b = \frac{1}{64}\). Applying the property helps in isolating \(x\) more easily.

Simplifying Radicals

Simplifying radicals is an essential step in solving equations involving square roots. The goal is to express the radical term as simply as possible. For square roots of fractions, like \[ \sqrt{\frac{1}{64}} \], we simplify by taking the square root of the numerator and the denominator separately. So, \[ \sqrt{\frac{1}{64}} = \frac{\sqrt{1}}{\sqrt{64}} = \frac{1}{8} \]. It’s important to practice simplifying different forms of radicals to become familiar with these techniques. This simplification makes subsequent algebraic manipulations more straightforward and the results neater.

Algebraic Solutions

Once the radical is simplified, we are left with an equation that is easier to solve algebraically. In our case, after simplifying \[ \sqrt{\frac{1}{64}} \] to \( \frac{1}{8} \), we have the equations \[ x - \frac{1}{8} = \pm \frac{1}{8} \]. We then split this into two separate equations: \[ x - \frac{1}{8} = \frac{1}{8} \] and \[ x - \frac{1}{8} = -\frac{1}{8} \]. Solving each one independently, we get \[ x = \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \] and \[ x = \frac{1}{8} - \frac{1}{8} = 0 \]. Always solve both parts to find all potential solutions.

Quadratic Equations

Quadratic equations come in the form \[ ax^2 + bx + c = 0 \]. They often require techniques like factoring, completing the square, or using the quadratic formula to solve. The exercise above demonstrated a specific method using the square root property for equations resembling \[ (x - a)^2 = b \]. Recognizing patterns in quadratic equations helps in choosing the most effective solving method. Overall, practicing varied forms of quadratic equations increases skill and confidence in solving them.