Problem 124 - 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 \(0 .\) $$s=s_{0}+g t^{2}+k \quad \text { for } t$$

Step-by-Step Solution

Answer

\(t = \pm \sqrt{\frac{s - s_0 - k}{g}}\)

Step 1: Isolate the quadratic term

We start with the formula \(s = s_0 + g t^2 + k\). We need to solve for \(t\), so first, we'll isolate the \(g t^2\) term by subtracting \(s_0\) and \(k\) from both sides of the equation. This gives us:\[s - s_0 - k = g t^2\]

Step 2: Divide by the coefficient of the quadratic term

To solve for \(t\), we need to get rid of the coefficient \(g\) in front of \(t^2\). Divide both sides by \(g\) to get:\[\frac{s - s_0 - k}{g} = t^2\]

Step 3: Solve for the variable by taking the square root

To isolate \(t\), take the square root of both sides of the equation. Remember that taking a square root introduces a \(\pm\) symbol:\[t = \pm \sqrt{\frac{s - s_0 - k}{g}}\]

Key Concepts

Isolate quadratic term

In solving quadratic equations, the foremost task is isolating the quadratic term. This process involves rearranging the equation to bring the quadratic component to one side. Picture this like untangling a thread—your goal is to separate the specific part you need to work with. Starting with the equation:\[s = s_0 + g t^2 + k\]and needing to solve for \( t \), our job is clear: isolate \( g t^2 \). We do this by subtracting \( s_0 \) and \( k \) from both sides:

• Subtract \( s_0 \) from both sides: \( s - s_0 = g t^2 + k \)
• Then, subtract \( k \) from both sides: \( s - s_0 - k = g t^2 \)

This procedure sets up the equation so the quadratic term \( g t^2 \) stands alone on one side, ready for the next step.

Solve for variable

Once the quadratic term is isolated, our next task is to solve for the variable—in this case, \( t \). This involves removing the coefficient in front of the variable. The equation you should have is:\[s - s_0 - k = g t^2\]Notice the \( g \) attached to \( t^2 \)? It's the coefficient, and it needs to be addressed. The simplest method is division:

• Divide every part of the equation by \( g \) to remove the coefficient on \( t^2 \).
• This gives you \( \frac{s - s_0 - k}{g} = t^2 \).

By dividing both sides by \( g \), we ensure that \( t^2 \) is now isolated, making the equation easier to manage for the subsequent steps.

Square root method

After solving for the variable up to a square term, we bring in the square root method. This method is about "unwrapping" the squared term.To finalize solving for \( t \), take the square root of both sides of the equation:\[ t = \pm \sqrt{\frac{s - s_0 - k}{g}}\]Why take the square root? Because it neutralizes the square on \( t \). The square root operation is the exact opposite of squaring a number, helping us to retrieve \( t \) itself, rather than \( t^2 \). Don't forget: taking a square root of both sides isn't just a mechanical step; it's the transition from \( t^2 \) back to \( t \).

Introduce ± symbol

An essential aspect of using the square root method is understanding the role of the \( \pm \) symbol. Whenever you take the square root of both sides of an equation, the \( \pm \) is introduced.Here's why: Squaring a negative and a positive number both result in the same positive number. Thus, when you reverse this operation using a root, both the positive and negative solutions must be acknowledged.Suppose someone only considers the positive case, they risk missing a perfectly valid solution.

• Positive root: \( \sqrt{x} \)
• Negative root: \( -\sqrt{x} \)

Hence, when you see \( \pm \sqrt{x} \), know it's there to ensure all possible values of \( t \) are represented. It enhances accuracy in solving equations and is a key step in quadratic problem-solving!