Problem 127 - A Graphical Approach to Precalculus with Limits

Question

Solve each formula for the indicated variable. Leave \(\pm\) in answers when applicable. Assume that no denominators are 0 $$P=\frac{E^{2} R}{(r+R)^{2}} \text { for } R$$

Step-by-Step Solution

Answer

\(R = \frac{E^2 - 2Pr \pm \sqrt{(2Pr - E^2)^2 - 4P^2r^2}}{2P}\)

Step 1: Clear the fraction

To remove the fraction, start by multiplying both sides of the equation by \((r+R)^{2}\) to get:\[P \cdot (r+R)^{2} = E^{2}R\]

Step 2: Expand and simplify

Expand \((r+R)^{2}\) to \(r^2 + 2rR + R^2\). Substitute back into the equation:\[P \cdot (r^2 + 2rR + R^2) = E^{2}R\] Distribute \(P\) on the left side:\[Pr^2 + 2PrR + PR^2 = E^2R\]

Step 3: Rearrange terms

Rearrange the equation to bring all terms involving \(R\) to one side:\[PR^2 + 2PrR - E^2R + Pr^2 = 0\]This simplifies to:\[PR^2 + (2Pr - E^2)R + Pr^2 = 0\]

Step 4: Use the quadratic formula

Identify this as a quadratic in terms of \(R\), where \(A = P\), \(B = 2Pr - E^2\), and \(C = Pr^2\). Use the quadratic formula \(R = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\):\[R = \frac{-(2Pr - E^2) \pm \sqrt{(2Pr - E^2)^2 - 4P \cdot Pr^2}}{2P}\]

Step 5: Simplify the formula

Simplify the expression inside the square root and calculate. The quadratic formula gives:\[R = \frac{E^2 - 2Pr \pm \sqrt{(2Pr - E^2)^2 - 4P^2r^2}}{2P}\]This is the solution for \(R\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in algebra for solving equations of the form \( ax^2 + bx + c = 0 \). It allows us to find the values of \(x\) that satisfy the equation. The formula is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \(a\), \(b\), and \(c\) are constants that are part of the quadratic equation. The symbol \( \pm \) indicates that there can be two solutions: one for the plus and one for the minus.
This formula is essential when working with quadratic equations, especially when factoring is complex or impossible. It utilizes the discriminant, \( b^2 - 4ac \), to determine the nature of the roots.

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, there is exactly one real solution (a repeated root).
• If the discriminant is negative, the equation has no real solutions, only complex ones.

In the original exercise, the quadratic formula helps us solve for \( R \), highlighting its utility in variable isolation.

Solving Equations

Solving equations involves finding the value of a variable that makes the equation true. This process can include various techniques such as simplifying expressions, combining like terms, and applying operations to both sides of the equation.
In the context of the original exercise, solving the equation for \( R \) requires us to manage both algebraic manipulation and understanding of quadratic structure. By multiplying both sides by \((r+R)^2\), we eliminate the fraction and create a standard form that enables further simplification and rearrangement.
The steps involve:

• Clearing fractions to simplify the equation.
• Rearranging terms to bring like terms together, preparing for applying the quadratic formula.
• Identifying the quadratic equation's components \(A\), \(B\), and \(C\) for formula application.

Each step systematically brings us closer to isolating the desired variable, which is a common goal when solving equations.

Variable Isolation

Variable isolation means rearranging an equation so that a specific variable becomes the subject of the formula. The goal is to express the variable entirely on one side of the equation, independent of the others.
Consider the example from the exercise: we aim to isolate \(R\), navigating algebraic complexities to solve the equation exclusively for this variable. This often requires moving terms around and performing operations such as addition, subtraction, multiplication, or division.
When attempting to isolate \(R\), we followed these steps:

• Removed the fraction by multiplying both sides by the denominator.
• Expanded and rearranged the resulting expression to form a quadratic equation.
• Utilized the quadratic formula, which provided the solution(s) for \(R\).

Just like solving equations generally involves finding the truth value of a variable, isolation focuses on one variable, indicating its dependency explicitly from the others, a critical skill in algebraic manipulation.