Problem 21 - Algebra 2 and Trigonometry

Question

In \(18-23,\) solve for the variable in each equation. Express the solution to the nearest hundredth. $$ 3 z^{3}+2=27 $$

Step-by-Step Solution

Answer

The solution is approximately \( z \approx 2.03 \).

Step 1: Isolate the cubic term

Start by subtracting 2 from both sides of the equation to isolate the cubic term on one side.\[ 3z^3 + 2 - 2 = 27 - 2 \] This simplifies to:\[ 3z^3 = 25 \]

Step 2: Solve for the cubic term

Divide both sides of the equation by 3 to solve for the cubic term:\[ z^3 = \frac{25}{3} \]

Step 3: Extract the cubic root

To solve for \( z \), take the cube root of both sides of the equation:\[ z = \sqrt[3]{\frac{25}{3}} \]

Step 4: Calculate and approximate the solution

Using a calculator, evaluate the cube root to find \( z \):\[ z \approx \sqrt[3]{8.3333} \approx 2.03008 \]

Step 5: Round the solution to the nearest hundredth

Round the calculated value of \( z \) to the nearest hundredth to find the final solution:\[ z \approx 2.03 \]

Key Concepts

Understanding the Cube Root

A cube root is essentially the opposite of cubing a number. If you cube a number, like 3 (which is 3 multiplied by itself twice more: 3 x 3 x 3), you get 27. Similarly, the cube root of 27 is 3. This operation is vital when dealing with cubic equations, as it helps us find a base number that, when cubed, will return the original value. In the exercise, we found the cube root of \( \frac{25}{3} \), or roughly 8.3333, which gives us the solution for \( z \). This process of finding the cube root allows us to solve equations where the variable is raised to the power of three. It's crucial in breaking down complex equations and moving forward in the problem-solving process.

Mastering Rounding Decimals

Rounding decimals helps simplify a number while keeping its value close to what it was. This is particularly useful when dealing with long decimal points in calculations where an exact figure isn't critical. In mathematics, rounding is often done to a specific decimal place, such as the nearest whole number, tenth, or hundredth.

• To round a number to the nearest hundredth, you look at the third decimal place.
• If that digit is 5 or more, you round the second decimal place up by one.
• If it is less than 5, you keep the second decimal place as it is.

For example, the number 2.03008 rounds to 2.03 because the third decimal, 0, is less than 5. Understanding how to round correctly is useful for presenting clearer, more readable solutions.

Approaching Equation Solving

Equation solving is the method by which we find an unknown value, typically represented by a variable such as \( z \). To solve an equation means to determine the value that makes the equation true. In the case of the given exercise, we worked through a series of logical steps to isolate \( z \), starting with simplifying the equation by removing constants and coefficients.

• First, isolate the term involving the variable: Subtract 2 from both sides so that \( 3z^3 \) stands alone.
• Next, divide both sides by 3 to solve for \( z^3 \).
• Finally, take the cube root of both sides to solve for \( z \).

This systematic approach to solving equations ensures that all operations done to one side of the equation are balanced with the other side, ultimately leading to the value of the variable.

Grasping the Concept of Approximation

Approximation is used in mathematics when an exact number is difficult or unnecessary to use. Instead, we find a number that is close enough to the correct answer for practical purposes. This is particularly important when the exact answer involves irrational numbers or repeating decimals. Using approximations, we analyzed the cube root of 8.3333 and found that it came out to an approximate 2.03008. This isn't an exact value, but it's sufficiently close for most practical applications in mathematics. Why is approximation important? It allows mathematicians and students alike to work with manageable and understandable numbers, while still capturing the essence of the numerical relationships they represent. This simplification is vital across multiple disciplines, from engineering to finance, where exact calculations could be unnecessarily complicated.