Problem 68 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ \left(x-\frac{1}{5}\right)^{2}=\frac{16}{25} $$

Step-by-Step Solution

Answer

x = 1 or x = -\(\frac{3}{5}\)

Step 1: Understand the Square Root Property

The square root property states that if \(a^2 = b\), then \(a = \pm \sqrt{b}\). Apply this property to the given equation.

Step 2: Take the Square Root of Both Sides

Start with the given equation: \(\left(x-\frac{1}{5}\right)^{2}=\frac{16}{25}\). Apply the square root property to both sides to get: \(x-\frac{1}{5} = \pm \sqrt{\frac{16}{25}}\).

Step 3: Simplify the Radical

Calculate the square root of \(\frac{16}{25}\). This equals \(\sqrt{\frac{16}{25}} = \frac{\sqrt{16}}{\sqrt{25}} = \frac{4}{5}\). Substitute this back into the equation: \(x-\frac{1}{5} = \pm \frac{4}{5}\).

Step 4: Solve for x

Consider the two cases separately:Case 1: \(x - \frac{1}{5} = \frac{4}{5}\).Add \(\frac{1}{5}\) to both sides: \(x = \frac{4}{5} + \frac{1}{5} = \frac{5}{5} = 1\).Case 2: \(x - \frac{1}{5} = -\frac{4}{5}\).Add \(\frac{1}{5}\) to both sides: \(x = -\frac{4}{5} + \frac{1}{5} = -\frac{3}{5}\).

Step 5: State the Solution

The solutions to the equation are \(x = 1\) and \(x = -\frac{3}{5}\).

Key Concepts

Solving Equations

Solving equations is a fundamental process in algebra that involves finding the value of the variable that makes an equation true. For instance, when given an equation like \(\frac{16}{25}\), you need to apply principles like the square root property to isolate and solve for the variable. In our example, it involves understanding how to deal with squares and square roots. By isolating the variable and performing inverse operations—like adding or subtracting terms from both sides—you can find the values that satisfy the equation. This method of systematic steps helps break down complex problems into manageable parts.

Simplifying Radicals

Simplifying radicals helps make equations less complicated and easier to solve. A radical expression or square root can be simplified by breaking it into smaller parts. For example, simplifying \(\frac{\root16}{25}\) involves finding the square roots of the numerator and denominator separately: \(\frac{4}{5}\). Breaking down the radicals step-by-step ensures that the solutions are as simple as possible, making it easier to solve the remaining steps of the equation.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They can often be solved using the square root property when they are formatted correctly. For example, in our exercise, we had \(\left(x - \frac{1}{5}\right)^2 = \frac{16}{25}\). By taking the square root of both sides, you reduce the quadratic equation to a simpler form. This approach is particularly helpful when the quadratic equation is already factored into a perfect square, allowing for the direct application of the square root property.

Algebraic Solutions

Algebraic solutions involve the steps and operations performed to solve algebraic equations. When working with our equation, \(\left(x - \frac{1}{5}\right)^2 = \frac{16}{25}\) requires the following:

• Applying the square root property
• Simplifying radicals
• Isolating the variable

. Each step adheres to algebraic principles to ensure that the solutions are accurate. Following these algebraic steps, we find that the solutions to our problem are \(x = 1\) and \(x = -\frac{3}{5}\). Understanding and applying these techniques are fundamental for accurately solving any algebraic equation.