Problem 8 - Elementary Algebra

Question

Solve the following equations, if possible. $$ r^{2}-49=0 $$

Step-by-Step Solution

Answer

Answer: The possible values of r are 7 and -7.

Step 1: Identify the quadratic equation

We are given the equation \(r^{2}-49=0\).

Step 2: Isolate \(r^2\)

To isolate \(r^2\), we need to add 49 to both sides of the equation: $$r^2 = 49$$

Step 3: Apply the square root property

To solve for \(r\), we need to take the square root of both sides of the equation. Remember that when taking the square root of both sides, we get two possible solutions: $$\sqrt{r^2} = \pm\sqrt{49}$$

Step 4: Calculate the square root

Now, we need to find the square root of 49: $$r = \pm\sqrt{49}$$ $$r = \pm7$$

Step 5: Write the solutions

We now have two possible solutions for \(r\): \(r = 7\) and \(r = -7\).

Key Concepts

Square Root Property

The square root property is a helpful tool when solving quadratic equations of the form \( x^2 = k \). This property suggests that if \( x^2 = k \), then \( x \) can be either the positive or negative square root of \( k \). In mathematical terms, this is expressed as:

• \( x = \pm \sqrt{k} \)

When you apply the square root property, it’s crucial to remember you always end up with two possible solutions. Consider our problem \( r^2 = 49 \). Applying the square root property, we take the square root of both sides:

• \( \sqrt{r^2} = \sqrt{49} \)

This results in \( r = \pm 7 \), providing both potential values for \( r \). Don't forget the \( \pm \) symbol, as this is what acknowledges both solutions.

Isolating the Variable

Before using the square root property, it's crucial to isolate the variable squared on one side of the equation. This process makes solving simpler and clearer by having a clear target to apply further operations. In our example, the original equation is \( r^2 - 49 = 0 \). You aim to have \( r^2 \) by itself on one side. To isolate \( r^2 \), you need to eliminate the constant number on its side.

• Here, add 49 to both sides: \( r^2 - 49 + 49 = 0 + 49 \)
• The equation becomes: \( r^2 = 49 \)

This step ensures \( r^2 \) is isolated, allowing you to apply further methods like the square root property efficiently.

Positive and Negative Solutions

When dealing with equations that involve squares, we often encounter two solutions: a positive and a negative one. This stems from the fundamental property of squaring where both positive and negative numbers, when squared, give a positive outcome. Let’s consider the number 7:

• \( 7^2 = 49 \)
• Also, \( (-7)^2 = 49 \)

Therefore, when solving \( r^2 = 49 \), we acknowledge two solutions: \( r = 7 \) and \( r = -7 \). Understanding that quadratic equations can yield both positive and negative solutions is key to taking full advantage of the square root property and fully solving the equation.

Basic Algebra

Basic Algebra involves foundational techniques that are necessary in solving equations. Understanding these techniques is crucial for progressing to more advanced mathematical concepts. In solving the equation \( r^2 - 49 = 0 \), we first apply a basic algebraic operation: adding 49 to both sides to isolate \( r^2 \). This step reflects the principle of maintaining equality by performing the same operation on both sides of an equation. Once isolated, the equation becomes \( r^2 = 49 \), ready for the application of the square root property. This process showcases basic algebraic concepts, like balancing equations, which are foundational for problem-solving in mathematics. Mastery of these basic skills is essential for tackling more complex algebraic equations.