Problem 14 - Intermediate Algebra

Question

Complete each solution. Solve \(8^{2}+4^{2}=c^{2}\) and assume \(c>0 .\) Approximate to two decimal places. $$\square+16=c^{2}$$ $$\quad\quad\quad\quad=\square{c}^{2}$$ $$\quad\sqrt{\square}=$$ $$\quad\sqrt\square{\cdot 5}=c$$ $$\quad\quad\sqrt{\square5}=c$$ $$c\quad \square\quad 8.94$$

Step-by-Step Solution

Answer

\(c \approx 8.94\), as \(\sqrt{80} \approx 8.94\).

Step 1: Solve for Missing Square Term

Begin by solving for the first missing square in the given problem, which is the square of 8. \(8^2 = 64\). Replace the question mark with 64 in the equation: \(64 + 16 = c^2\).

Step 2: Solve for the Value of \(c^2\)

Add the values 64 and 16 to find \(c^2\). \(64 + 16 = 80\). Therefore, \(c^2 = 80\). Substitute 80 into the equation, making it: \(c^2 = 80\).

Step 3: Calculate \(c\) by Taking the Square Root

Take the square root of both sides of the equation to solve for \(c\). \(c = \sqrt{80}\). Simplify \(\sqrt{80}\), which is approximately \(8.94\).

Step 4: Fill in the Missing Evaluation Steps

Use the approximate result from the square root. \(c \approx 8.94\). Therefore, complete the last statement with this value.

Key Concepts

Solving Equations

When solving equations, especially involving the Pythagorean Theorem, it's crucial to understand how to isolate the variable you're solving for. Here, we are dealing with the equation \(8^2 + 4^2 = c^2\). This is a common setup derived from the Pythagorean Theorem which states that, for a right-angled triangle, the square of the hypotenuse (\(c\)) is equal to the sum of the squares of the other two sides (in this case, 8 and 4).

To solve for \(c^2\), begin by calculating each square:

• 8 squared \((8^2)\) equals 64
• 4 squared \((4^2)\) equals 16

Next, add these two results to get \(c^2\) :

• 64 plus 16 equals 80

Therefore, \(c^2 = 80\). This equation tells you that the square of the hypotenuse \(c\) must equal 80. Our task is to solve for \(c\). This is achieved by finding the square root of \(c^2\). A clear understanding of such setups will greatly enhance your problem-solving skills.

Square Roots

Understanding square roots is essential, as they are a foundational concept in math, especially in geometry and algebra. In the context of the Pythagorean Theorem \(c^2 = 80\), we solve for \(c\) by taking the square root of both sides of the equation. This can be represented as:

• \(c = \sqrt{80}\)

Taking the square root is the inverse operation of squaring a number. Here, we deal with an approximate value because 80 is not a perfect square.

We recognize that the square root simplifies by breaking \(80\) down into its factors. Thus:

• \(\sqrt{80} = \sqrt{16 \cdot 5} = \sqrt{16} \times \sqrt{5}\)
• \(\sqrt{16} = 4\)

Therefore, \(\sqrt{80}\) simplifies to \(4 \times \sqrt{5}\). Approximating \(\sqrt{5}\) gives a more accessible numeric form. Knowing how to simplify square roots is a valuable tool in solving more complex equations and makes approximations easier to handle.

Approximation Methods

Approximating a square root when it isn't a perfect square is an important skill. Once we have simplified \(\sqrt{80} = 4 \times \sqrt{5}\), the next step is to approximate \(\sqrt{5}\) since it does not result in a precise whole number.

One common method for approximating square roots is using a calculator, which is practical for quick results. The square root of 5 is roughly 2.236. Therefore:

• \(c = 4 \times 2.236\)

Calculating this product gives us 8.944, which we round to 8.94 for precision to two decimal places. Using approximations allows us to work with more accurate, manageable numbers in real-world applications and further mathematics.