Problem 39 - Elementary Algebra

Question

Solve using the Square Root Property. \(\left(p-\frac{1}{3}\right)^{2}=\frac{7}{9}\)

Step-by-Step Solution

Answer

p = 1 and 0.

Step 1 - Recognize the Square Root Property

The Square Root Property states that if can take the square root of both sides using two signs: plus and minus.

Step 2 - Take the Square Root of Both Sides

Take the square root of both sides of the equation: .

Step 3 - Simplify the Equation

Simplify the equation by using the square root of the fraction . The equation simplifies to: using: both positive and negative roots.

Step 4 - Solve for p

To solve for , add both sides for two solutions: and .

Step 5 - Write the Final Answer

Combining both solutions, the final answers are: and .

Key Concepts

Solving Quadratic Equations

The Square Root Property is a valuable tool for solving quadratic equations. It is especially helpful when the quadratic equation is in the form \( a^2 = b \). The property indicates that if \( x^2 = k \), then \( x = \pm\sqrt{k} \).

Let's apply this to our given exercise: \( \left(p-\frac{1}{3}\right)^{2} = \frac{7}{9} \).

• Step 1: Identify that the equation is in a form which allows the application of the Square Root Property.

By applying the Square Root Property, we can express it as: \( p-\frac{1}{3} = \pm\sqrt{\frac{7}{9}} \).

This step simplifies the entire solving process by reducing the equation to two simpler linear equations. By solving both of these equations, we will get two potential solutions for \( p \).

Simplifying Square Roots

Next, we need to simplify the square root expression: \( \sqrt{\frac{7}{9}} \).

• Step 2: Simplify the Square Root.

Breaking it down, we first find the square root of the numerator and the denominator separately: \( \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{\sqrt{9}} \).

The square root of 9 is a known value, which is 3. Hence, we get: \( \frac{\sqrt{7}}{3} \).

Now, we can rewrite the equation from our previous step as: \( p - \frac{1}{3} = \pm \frac{\sqrt{7}}{3} \). This gives us two linear equations: \( p - \frac{1}{3} = \frac{\sqrt{7}}{3} \) and \( p - \frac{1}{3} = -\frac{\sqrt{7}}{3} \).

Finding Solutions to Equations

Finally, we solve for \( p \) in these linear equations.

• Step 3: Solve for \( p \).

For the first equation: \( p - \frac{1}{3} = \frac{\sqrt{7}}{3} \), we add \( \frac{1}{3} \) to both sides: \( p = \frac{\sqrt{7}}{3} + \frac{1}{3} = \frac{\sqrt{7} + 1}{3} \).

For the second equation: \( p - \frac{1}{3} = -\frac{\sqrt{7}}{3} \), we also add \( \frac{1}{3} \) to both sides: \( p = -\frac{\sqrt{7}}{3} + \frac{1}{3} = \frac{-\sqrt{7} + 1}{3} \).

Combining both solutions, we get: \( p = \frac{\sqrt{7} + 1}{3} \) and \( p = \frac{-\sqrt{7} + 1}{3} \). Therefore, the final solutions to the equation are \( p = \frac{\sqrt{7} + 1}{3} \) and \( p = \frac{-\sqrt{7} + 1}{3} \).

By following these detailed steps, any quadratic equation that can be simplified using the Square Root Property becomes manageable. Understanding and applying these concepts is the key to mastering such problems in algebra.