Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(39-56:\) Find the solutions of the equation. $$ x^{3}+125=0 $$

Step-by-Step Solution

Answer

The solution is \(x = -5\).

Step 1: Isolate the Cubic Term

The given equation is \(x^3 + 125 = 0\). Start by isolating the cubic term by subtracting 125 from both sides of the equation to obtain \(x^3 = -125\).

Step 2: Solve for x by Taking the Cube Root

To solve for \(x\), take the cube root of both sides. Recall that the cube root of a negative number is also negative, so:\(x = \sqrt[3]{-125}\).

Step 3: Calculate the Cube Root

Calculate the cube root of \(-125\). Knowing that \(-125 = -5^3\), we have \(\sqrt[3]{-125} = -5\).

Step 4: Verify the Solution

Substitute \(x = -5\) back into the original equation to ensure it satisfies the equation: \((-5)^3 + 125 = -125 + 125 = 0\), which is true. Thus, \(x = -5\) is a solution.

Key Concepts

Cubic Roots

Cubic roots are the values that, when multiplied by themselves twice, give the original number. For instance, if you want the cubic root of 8, you are looking for a number that gives 8 when cubed. In mathematical notation, this is expressed as \( x^3 = 8 \), and the solution is \( x = 2 \) because \( 2^3 = 8 \). In this exercise, we encountered the cubic root of \(-125\).

Unlike square roots, cubic roots can yield both negative and positive results depending on the original number's sign. The cube root of a negative number is simply the negative of the cube root of its positive counterpart.

• For example, \( \sqrt[3]{-125} = -5 \), as \( -5 \times -5 \times -5 = -125 \).
• Cube roots are denoted as \( \sqrt[3]{x} \), where \( x \) can be any real number.

Understanding this concept is crucial in solving equations involving cubic terms.

Solutions of Equations

To solve equations, especially polynomial equations like the one in this exercise, we often aim to isolate the variable term first. Let's consider the equation from our exercise: \( x^3 + 125 = 0 \). Our first task is to get \( x^3 \) by itself:

• Subtract 125 from both sides of the equation to isolate \( x^3 \): \( x^3 = -125 \).

Once the variable is isolated, you determine the roots. Here, since the equation is simple, the primary goal is to find the single cube root solution. Because these types of equations can have up to three solutions, identifying real and imaginary roots is important. However, for this equation, the solution is a real number.

When solving polynomial equations, look for the number of potential roots and remember that their types (real or complex) depend on the polynomial's degree and the discriminant.

Verification of Solutions

Verification is a crucial step in solving equations as it confirms the accuracy of your solution. Let's take the provided solution, \( x = -5 \), and plug it back into the original equation \( x^3 + 125 = 0 \):

• Substitute \( x = -5 \) into the equation: \( (-5)^3 + 125 \).
• Calculate \( (-5)^3 = -125 \).
• Adding 125 yields: \( -125 + 125 = 0 \).

Because the equation equals zero, it verifies that \( x = -5 \) is indeed a correct solution. If you find different values upon substitution, it's an indication that either the solution or your initial steps might need revisiting. Verification ensures that all calculations were performed correctly and instills confidence in the solution's validity.