Question

Change the subject of the formula L = v 4kt - p to k.

Answer 1

Answer:

$\boxed{k = \frac{L^2 + p}{4t}}$

General Formulas and Concepts:

Algebra I

Basic Equality Properties

1. Multiplication Property of Equality
2. Division Property of Equality
3. Addition Property of Equality
4. Subtraction Property of Equality

Step-by-step explanation:

Step 1: Define

Identify given.

$\displaystyle L = \sqrt{4kt - p}$

Step 2: Solve for k
We can use equality properties to help us rewrite the equation to get k as our subject:

Let's first square both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\\end{aligned}$

Next, add p to both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\\end{aligned}$

Next, divide 4t by both sides:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow L^2 = \big( \sqrt{4kt - p} \big) ^2 \\& \rightarrow L^2 = 4kt - p \\& \rightarrow L^2 + p = 4kt \\& \rightarrow \frac{L^2 + p}{4t} = k \\\end{aligned}$

We can rewrite the new equation by swapping sides to obtain our final expression:

$\displaystyle\begin{aligned}L = \sqrt{4kt - p} & \rightarrow \boxed{k = \frac{L^2 + p}{4t}}\end{aligned}$

∴ we have changed the subject of the formula.

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Topic: Algebra I