Problem 16 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ \sqrt[3]{6-s^{2}}+5=0 $$

Step-by-Step Solution

Answer

\(s = \sqrt{131}\) or \(s = -\sqrt{131}\)

Step 1: Isolate the Radicand

The given equation is \( \sqrt[3]{6-s^{2}}+5=0 \). To start, we need to isolate the cube root. Subtract 5 from both sides of the equation to get: \( \sqrt[3]{6-s^2} = -5 \).

Step 2: Eliminate the Cube Root

Now that the cube root is isolated, we remove it by cubing both sides of the equation. This gives: \( 6-s^2 = (-5)^3 \). Calculating \((-5)^3\), we find \(-125\). So, the equation becomes \(6-s^2 = -125 \).

Step 3: Solve for \(s^2\)

Rearrange the equation to solve for \(s^2\) by subtracting 6 from both sides, getting \(-s^2 = -131 \). To find \(s^2\), multiply both sides by -1: \(s^2 = 131 \).

Step 4: Solve for \(s\)

To find \(s\), take the square root of both sides of the equation. Since \(s^2 = 131\), we get \(s = \pm \sqrt{131} \). This results in two potential solutions: \(s = \sqrt{131}\) and \(s = -\sqrt{131}\).

Key Concepts

Cube Root

Understanding the cube root is crucial when solving equations that contain terms with exponents or roots. The cube root of a number is a value that, when multiplied by itself three times, will give the original number. In mathematical terms, if you have a number \( x \), then \( \sqrt[3]{x} \) is the cube root.

This is particularly useful for solving equations where the term is trapped inside a cube root, as it allows you to "undo" the cubing process. For example, \( \sqrt[3]{27} = 3 \) because when 3 is cubed (\( 3 \times 3 \times 3 \)), it equals 27. By applying this understanding, you can remove the cube root in equations to find the solution for the variable involved.

Solving Equations

Solving equations is about finding the values of the unknowns that make the equation true. The process involves systematic steps including manipulating the equation to isolate terms and simplify expressions.

Methods for solving equations include:

• Rearranging terms to get the unknown variable on one side.
• Performing arithmetic operations to both sides of the equation to maintain balance.
• Applying inverse operations like squaring or taking roots when necessary.

A key principle is to perform the same operation on both sides of the equation to keep it balanced.

Isolating Variables

Isolating variables is an essential step in solving equations. It involves rearranging the equation to get the unknown variable by itself on one side. This step is crucial as it allows you to solve for the variable.

Here’s how you isolate a variable:

• Look at the equation and determine which term contains the variable you need to solve for.
• Use operations like addition, subtraction, multiplication, or division to move other terms to the opposite side of the equation.
• Ensure that any fractions or coefficients are cleared, if necessary, to leave the variable alone.

In the given problem, isolating \( s^2 \) was key to reaching the solution.

Radicand

The radicand is the number under the radical symbol, such as under a square or cube root. It’s important because manipulating the radicand can change the value of the root, directly affecting solutions.

• In the equation \( \sqrt[3]{6-s^2}+5=0 \), the entire expression \( 6-s^2 \) is the radicand.
• Isolating the radicand is necessary to eliminate the root, making it easier to solve the equation.
• Changes to the radicand will reflect on the result of taking the root.

Understanding the role of the radicand allows one to navigate through complex root-based equations efficiently.

Square Roots

Square roots are common operations in algebra and involve finding a number that, when multiplied by itself, equals the original number. For example, the square root of 25 is 5 because \( 5 \times 5 = 25 \).

When dealing with equations, taking a square root can help resolve squared terms:

• To remove a square, apply the square root to both sides of an equation.
• Remember that taking a square root results in both positive and negative roots, denoted by \( \pm \).
• This concept is crucial when encountering terms like \( s^2 \) in an equation, helping find potential solutions.

Square roots can lead to two possible solutions due to the nature of squaring numbers, as shown in solving for \( s^2 = 131 \).