Problem 66 - Beginning and Intermediate Algebra

Question

Solve using the square root property. Simplify all radicals. $$ \left(x-\frac{1}{9}\right)^{2}=\frac{1}{81} $$

Step-by-Step Solution

Answer

x = \frac{2}{9}, 0

Step 1 - Apply the Square Root Property

To solve the equation \[ \left(x - \frac{1}{9} \right)^{2} = \frac{1}{81} \], apply the square root property by taking the square root of both sides of the equation. This gives: \[ x - \frac{1}{9} = \pm \sqrt{\frac{1}{81}} \]

Step 2 - Simplify the Radical

Next, simplify the square root. The square root of \( \frac{1}{81} \) is \( \frac{1}{9} \). Thus, we have: \[ x - \frac{1}{9} = \pm \frac{1}{9} \]

Step 3 - Solve for x

Solve the equation for \( x \) by splitting into two cases: \[ \text{Case 1: } x - \frac{1}{9} = \frac{1}{9} \] \[ \text{Add } \frac{1}{9} \text{ to both sides: } x = \frac{1}{9} + \frac{1}{9} = \frac{2}{9} \] \[ \text{Case 2: } x - \frac{1}{9} = - \frac{1}{9} \] \[ \text{Add } \frac{1}{9} \text{ to both sides: } x = - \frac{1}{9} + \frac{1}{9} = 0 \]

Step 4 - Conclusion

The solutions to the equation are \( x = \frac{2}{9} \) and \( x = 0 \).

Key Concepts

Quadratic Equations

Quadratic equations are an essential part of algebra. They are characterized by their highest power, which is always two. The general form of a quadratic equation is: \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are coefficients and \(x\) represents an unknown variable.
Quadratic equations can be solved using several methods. Common techniques include:

• Factoring
• Completing the square
• Quadratic formula
• Square root property

For our specific problem, we use the square root property, which is effective when our equation is already in the form \((x - p)^2 = q\). Let's explore this method further!

Radical Simplification

Radical simplification is a process where we simplify expressions under a square root (or any radical). This step is crucial to make solving equations easier.
In our example, we start with the equation: \[ (x - \frac{1}{9})^2 = \frac{1}{81} \]The next step is to apply the square root property, taking the square root of both sides: \[ x - \frac{1}{9} = \frac{1}{9} \text{ or } x - \frac{1}{9} = -\frac{1}{9} \]Now, we simplify: The square root of \(\frac{1}{81}\) is \( \frac{1}{9} \), because: \[ \frac{1}{81} = \frac{1^2}{9^2} \rightarrow \text{so} \sqrt{\frac{1}{81}} = \frac{1}{9} \]Simplifying radicals helps us get clean, manageable numbers to work with.

Algebraic Solutions

Algebraic solutions involve isolating the variable we're solving for, using proper algebra rules.
Returning to our simplified expression: \[ x - \frac{1}{9} = \frac{1}{9} \text{ or } x - \frac{1}{9} = -\frac{1}{9} \]We separate this into two cases.

• Case 1
  \[ x - \frac{1}{9} = \frac{1}{9} \rightarrow x = \frac{1}{9} + \frac{1}{9} = \frac{2}{9} \]
• Case 2
  \[ x - \frac{1}{9} = -\frac{1}{9} \rightarrow x = -\frac{1}{9} + \frac{1}{9} = 0 \]

So, the solutions to our quadratic equation are: \[ x = \frac{2}{9} \text{ and } x = 0 \]