Problem 69 - Elementary and Intermediate Algebra

Question

Solve each equation for the specified variable or expression. $$ L_{A}=L_{B} \sqrt{1-\frac{v^{2}}{c^{2}}} \text { for } v^{2} $$

Step-by-Step Solution

Answer

The solution is \( v^2 = c^2 \left( 1 - \left( \frac{L_{A}}{L_{B}} \right)^2 \right) \).

Step 1: Isolate the square root expression

We start with the given equation:\[ L_{A} = L_{B} \sqrt{1 - \frac{v^{2}}{c^{2}}} \]Our goal is to solve for \( v^2 \). First, divide both sides by \( L_B \) to isolate the square root expression:\[ \frac{L_{A}}{L_{B}} = \sqrt{1 - \frac{v^{2}}{c^{2}}} \]

Step 2: Eliminate the square root

Next, eliminate the square root by squaring both sides of the equation:\[ \left( \frac{L_{A}}{L_{B}} \right)^2 = 1 - \frac{v^{2}}{c^{2}} \]

Step 3: Rearrange to isolate \( v^2/c^2 \)

Re-arrange the equation to solve for \( \frac{v^2}{c^2} \):\[ \frac{v^2}{c^2} = 1 - \left( \frac{L_{A}}{L_{B}} \right)^2 \]

Step 4: Solve for \( v^2 \)

Finally, multiply both sides of the equation by \( c^2 \) to solve for \( v^2 \):\[ v^2 = c^2 \left( 1 - \left( \frac{L_{A}}{L_{B}} \right)^2 \right) \]

Key Concepts

Solving Equations

Solving equations is a fundamental skill in algebra, crucial for manipulating mathematical relationships. It involves finding the value of the variable that makes the equation true. When you have an equation, both sides are equal under specific conditions for the variable involved. In our exercise, the goal is to solve for \( v^2 \). We began by identifying the part of the equation that needs to be isolated or simplified.

• Identify the goal: Decide which variable or equation part you're solving for. Here, it's \( v^2 \).
• Perform operations equally: You must perform the same operation on both sides of the equation to maintain balance. For example, dividing both sides by \( L_B \).

By consistently applying algebraic principles, you move step by step towards isolating the target variable or expression. This problem-solving technique is widely applicable to various complex equations.

Square Roots

Square roots are used to determine what original number was squared to obtain the given number. In equations, they often appear within expressions that need simplifying. In our exercise, the equation root function was involved in the set up:

• Understanding square root: The square root of a number is a value which, when multiplied by itself, yields the original number. In this exercise, \( \sqrt{1 - \frac{v^2}{c^2}} \) needed to be simplified.
• Eliminating square roots: To solve our equation, we needed to eliminate the square root by squaring both sides. This reversed the operation, allowing further simplification.

Remember, when squaring both sides of an equation, you must consider both the principal (positive) and negative roots, though the problem's context might limit which is physically meaningful.

Isolation of Variables

Isolating variables involves rearranging an equation to get the expression or variable of interest alone on one side. This allows you to solve for its value directly.

• Balancing equations: Ensure each step taken to isolate a variable keeps the equation balanced, meaning operations must be applied equally to both sides.
• Step-by-step isolation: In our exercise, the key steps involved isolating \( \sqrt{1 - \frac{v^2}{c^2}} \), then \( \frac{v^2}{c^2} \), before finally arriving at \( v^2 \).

This process ultimately involves moving terms using addition, subtraction, multiplication, or division, respecting the operation order to untangle the variable from other numerical or variable expressions. These skills are pivotal for tackling more complex physics or engineering problems, where isolating terms is necessary for interpreting or predicting results.