Problem 116 - A Graphical Approach to College Algebra

Question

Solve each formula for the indicated variable. Leave \(\pm\) in answers when appropriate. Assume that no denominators are \(0 .\) $$\mathscr{A}=\pi r^{2} \quad \text { for } r$$

Step-by-Step Solution

Answer

\( r = \pm \sqrt{\frac{\mathscr{A}}{\pi}} \)

Step 1: Understand the Formula

We are given the formula for the area of a circle: \( \mathscr{A} = \pi r^2 \). We need to solve this equation for \( r \), which represents the radius of the circle.

Step 2: Isolate the Radius Term

Start by dividing both sides of the equation \( \mathscr{A} = \pi r^2 \) by \( \pi \) to isolate \( r^2 \) on one side of the equation. This gives us \( \frac{\mathscr{A}}{\pi} = r^2 \).

Step 3: Solve for Radius

To solve for \( r \), take the square root of both sides of the equation \( r^2 = \frac{\mathscr{A}}{\pi} \). This will yield the solution \( r = \pm \sqrt{\frac{\mathscr{A}}{\pi}} \), representing both positive and negative roots.

Step 4: Choose the Meaningful Solution

Since \( r \) represents the radius of a circle, which must be a non-negative value, we take only the positive root. Therefore, \( r = \sqrt{\frac{\mathscr{A}}{\pi}} \). However, if asked to leave \( \pm \) in answers when appropriate, note both roots: \( r = \pm \sqrt{\frac{\mathscr{A}}{\pi}} \).

Key Concepts

Area of a Circle

The area of a circle is a fundamental concept in geometry that describes the space contained within a circle's boundary. The formula for calculating the area uses the radius and the constant π (Pi), which is approximately 3.14159. The formula is:\[ \mathscr{A} = \pi r^2 \]Here, \( \mathscr{A} \) represents the area, and \( r \) is the radius. The radius is the distance from the center of the circle to any point on its edge.

This formula illustrates a key relationship: the area of a circle grows with the square of its radius. This means if you double the radius, the area multiplies by four! Understanding this formula is essential for a wide range of applications, from engineering to everyday problem-solving.

Quadratic Equations

Quadratic equations are mathematical expressions that equate a parabola, which is a type of curve, in algebra. They typically take the general form:\[ ax^2 + bx + c = 0 \]where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. Solving these equations involves finding the values of \( x \) that make the equation true. This is often done using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]The term under the square root, \( b^2 - 4ac \), is known as the discriminant, and it determines:

• Whether there are real or complex solutions
• How many solutions exist

Quadratic equations are very important and appear frequently in various fields like physics, economics, and biology. They describe situations such as projectile motion or object trajectories.

Solving for a Variable

Solving for a variable means isolating that variable on one side of the equation and expressing it in terms of the remaining quantities. This is often necessary in algebra when rearranging formulas or solving equations.

The process typically involves a few steps:

• Identify the variable: Determine which variable you need to isolate.
• Use inverse operations: Apply operations that undo the current ones, like addition with subtraction or multiplication with division.
• Simplification: Simplify the equation step by step until the variable is by itself on one side.

For instance, in the equation \( \mathscr{A} = \pi r^2 \), we isolated \( r \) by dividing by \( \pi \) and then taking the square root. Such solutions require careful attention to arithmetic rules and sometimes involve recognizing that multiple solutions (positive and negative) are possible, as shown by \( \pm \sqrt{\frac{\mathscr{A}}{\pi}} \).