Problem 53 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0 $$

Step-by-Step Solution

Answer

The real solution is \( x = -\frac{1}{2} \); \( x = 0 \) is not allowed.

Step 1: Simplify the Equation

We start by realizing that we have a common factor of \(\frac{1}{x}\) across all terms, which allows us to factor it out. The equation becomes:\[ \frac{1}{x}\left( \frac{1}{x^2} + \frac{4}{x} + 4 \right) = 0 \] This implies two possible solutions: either \( x = 0 \) or the expression inside the parenthesis equals zero.

Step 2: Solve for x when x ≠ 0

We need to find solutions for when the expression inside the parenthesis \[ \frac{1}{x^2} + \frac{4}{x} + 4 = 0 \] is equal to zero. This is a quadratic equation in terms of \( \frac{1}{x} \) if we substitute \( y = \frac{1}{x} \), then it transforms into:\[ y^2 + 4y + 4 = 0 \].

Step 3: Solve the Quadratic Equation

The quadratic equation \( y^2 + 4y + 4 = 0 \) can be solved using the quadratic formula:\[ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]where \( a = 1, b = 4, c = 4 \). Substituting these values gives:\[ y = \frac{{-4 \pm \sqrt{{16 - 16}}}}{2} = \frac{{-4}}{2} = -2 \].Therefore, \( y = -2 \).

Step 4: Substitute Back to Find x

Recall that \( y = \frac{1}{x} \). Since \( y = -2 \), this implies \( \frac{1}{x} = -2 \) or \( x = \frac{1}{-2} = -\frac{1}{2} \).

Step 5: Verify the Solution

We need to verify that this solution \( x = -\frac{1}{2} \) satisfies the original equation:Substitute \( x = -\frac{1}{2} \) back into the original equation:\[ \frac{1}{(-\frac{1}{2})^3} + \frac{4}{(-\frac{1}{2})^2} + \frac{4}{(-\frac{1}{2})} = -8 + 16 - 8 = 0 \].The original equation holds, confirming that \( x = -\frac{1}{2} \) is a valid solution.

Key Concepts

Factoring

Factoring is a fundamental process that involves simplifying expressions by breaking them down into products of simpler factors. In the context of rational equations, factoring can simplify the process of finding solutions by isolating potential points of interest, such as zeros of a function.

In our given problem, the equation \( \frac{1}{x^3} + \frac{4}{x^2} + \frac{4}{x} = 0 \), has a common factor of \( \frac{1}{x} \) in each term. This means we can simplify the equation by factoring out \( \frac{1}{x} \), leading to:

• \( \frac{1}{x} \left( \frac{1}{x^2} + \frac{4}{x} + 4 \right) = 0 \)

This gives us two conditions to consider:

• \( \frac{1}{x} = 0 \), which would imply \( x = 0 \) (though this solution is not valid as it would create undefined expressions)
• The quadratic part \( \frac{1}{x^2} + \frac{4}{x} + 4 = 0 \) must be zero

By focusing on the quadratic expression, we can proceed to solve for \( x \), avoiding undefined values by incorporating substitutions.

Quadratic Formula

The quadratic formula is a versatile tool that helps solve equations of the form \( ax^2 + bx + c = 0 \). In our exercise, after factoring, we're left with the equation \( \frac{1}{x^2} + \frac{4}{x} + 4 = 0 \), which is still in terms of \( x^{-1} \) rather than \( x \).

Substituting \( y = \frac{1}{x} \) transforms our equation into a standard quadratic form:

• \( y^2 + 4y + 4 = 0 \)

The coefficients here are \( a = 1 \), \( b = 4 \), and \( c = 4 \). By inputting these into the quadratic formula:

• \[ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
• \[ y = \frac{{-4 \pm \sqrt{{16 - 16}}}}{2} = \frac{{-4}}{2} = -2 \]

Thus, the solution to \( y = \frac{1}{x} = -2 \) corresponds to \( x = -\frac{1}{2} \). The quadratic formula helps to derive exact solutions rooted in algebraic simplicity.

Verification of Solutions

After solving an equation, it's crucial to verify that the derived solutions satisfy the original equation, especially when dealing with rational equations which might have restrictions.

For our solution \( x = -\frac{1}{2} \), we substitute it back into the original equation to ensure accuracy:

• Original: \( \frac{1}{x^3} + \frac{4}{x^2} + \frac{4}{x} = 0 \)
• Substitute: \( \frac{1}{\left(-\frac{1}{2}\right)^3} + \frac{4}{\left(-\frac{1}{2}\right)^2} + \frac{4}{\left(-\frac{1}{2}\right)} = -8 + 16 - 8 \)

Calculating gives zero, confirming the solution is correct. Verification is a key step as it assures that calculations do not merely satisfy transformed equations but also meet the initial conditions present in the problem.