Problem 54 - Prealgebra

Question

Solve each equation. Round to the nearest tenth, if necessary. $$168=2 p^{2}$$

Step-by-Step Solution

Answer

\( p \approx \pm 9.2 \)

Step 1: Isolate the Variable Term

To solve for \( p \), start by isolating the \( p^2 \) term. Divide both sides of the equation by 2: \[ \frac{168}{2} = \frac{2p^{2}}{2} \] This simplifies to:\[ 84 = p^{2} \]

Step 2: Solve for the Variable

Next, solve for \( p \) by taking the square root of both sides. Remember, when you take the square root of both sides, you consider both the positive and negative roots:\[ p = \pm \sqrt{84} \]

Step 3: Simplify the Square Root and Round

Calculate \( \sqrt{84} \) and round to the nearest tenth. Using a calculator, you find:\[ \sqrt{84} \approx 9.165 \]Rounding 9.165 to the nearest tenth gives 9.2, so:\[ p \approx \pm 9.2 \]

Key Concepts

Isolating the Variable

In the journey of solving quadratic equations, the first and crucial step is isolating the variable we want to solve for. Here, our focus is on the variable \( p \). To begin, identify the term that contains this variable—in this case, \( 2p^2 \). The goal is to have \( p^2 \) alone on one side of the equation.

This involves basic arithmetic operations such as addition, subtraction, multiplication, or division. In our exercise, we divide both sides by 2, which is the coefficient of \( p^2 \). This operation effectively isolates \( p^2 \) on the left side:
\[ 84 = p^{2} \]
This step is important because it prepares the equation for further operations, bringing us closer to finding the value of \( p \). Simplifying equations by isolating the variable term is a fundamental skill in algebra.

Square Root Method

After isolating the variable, the next logical step is to solve for \( p \) using the square root method. This method involves taking the square root of both sides of the equation to solve for the variable.

For the equation \( 84 = p^2 \), you take the square root of both sides to find \( p \):
\[ p = \pm \sqrt{84} \]
It's crucial to remember that when you deal with quadratic equations, each square root solution includes both its positive and negative values. Thus, you get two possible solutions for \( p \):

• \( p = +\sqrt{84} \)
• \( p = -\sqrt{84} \)

The square root method is a particularly straightforward approach in these situations and helps to find the potential roots of quadratic equations efficiently.

Rounding to the Nearest Tenth

After finding the possible values of \( p \) using the square root, it's essential to express the result practically, often by rounding. Here, we need to round to the nearest tenth.

You begin by calculating \( \sqrt{84} \) with a calculator, which results in approximately 9.165.
Now, focusing on the decimal part, you round 9.165 to the nearest tenth. This means evaluating the hundreds place (the second digit after the decimal):

• If it's 5 or more, round up.
• If it's less than 5, keep the tenths digit the same.

Since 9.165 rounds up due to the 6 in the hundredths, \( p \) becomes approximately \( 9.2 \).

Thus, the rounded solutions for \( p \) are:

• \( p \approx +9.2 \)
• \( p \approx -9.2 \)

Rounding ensures that answers are presented in a readable, standardized format.