Problem 15 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{x^{3}-4 x}{x^{2}+1}=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 0 \), \( x = 2 \), and \( x = -2 \).

Step 1: Simplify the Equation

The given equation is \( \frac{x^{3}-4x}{x^{2}+1}=0 \). For the fraction to be zero, the numerator must be zero. Hence, set the numerator equal to zero: \( x^{3} - 4x = 0 \).

Step 2: Factor the Numerator

Factor the expression \( x^{3} - 4x = 0 \). This can be factored as \( x(x^{2} - 4) = 0 \).

Step 3: Factor Further and Solve

The expression \( x^{2} - 4 \) can be factored further as \( (x - 2)(x + 2) \). Substitute this back into the equation: \( x(x - 2)(x + 2) = 0 \). This gives us three potential solutions by setting each factor equal to zero: \( x = 0 \), \( x = 2 \), and \( x = -2 \).

Step 4: Verify the Solutions

To verify, substitute each solution back into the original equation: For \( x = 0 \): \( \frac{0^{3} - 4(0)}{0^{2} + 1} = 0 \) is correct. For \( x = 2 \): \( \frac{2^{3} - 4(2)}{2^{2} + 1} = \frac{8 - 8}{4 + 1} = 0 \) is correct. For \( x = -2 \): \( \frac{(-2)^{3} - 4(-2)}{(-2)^{2} + 1} = \frac{-8 + 8}{4 + 1} = 0 \) is correct. All solutions satisfy the original equation.

Key Concepts

Factoring Equations

Factoring is a method used to simplify complex equations and find their solutions. It's like breaking down a big problem into smaller, more manageable parts. To solve the problem \( \frac{x^{3}-4x}{x^{2}+1}=0 \), we focus on the numerator: \( x^{3} - 4x \). Because a fraction equals zero only when its numerator is zero, we set \( x^{3} - 4x = 0 \).

This expression can be factored by pulling out the greatest common factor, which is \( x \). So, it becomes \( x(x^{2} - 4) = 0 \). From here, \( x^{2} - 4 \) is recognizable as a difference of squares and can be further factored to \( (x - 2)(x + 2) \).

Now the equation is \( x(x - 2)(x + 2) = 0 \). To find the solutions, set each factor equal to zero:

• \( x = 0 \)
• \( x - 2 = 0 \) leads to \( x = 2 \)
• \( x + 2 = 0 \) leads to \( x = -2 \)

Through factoring, we discovered three potential solutions.

Verifying Solutions

Verification of solutions confirms whether they actually solve the original equation. After finding potential solutions from the factors \( x = 0 \), \( x = 2 \), and \( x = -2 \), we must check if they satisfy the original equation \( \frac{x^{3}-4x}{x^{2}+1} = 0 \).

To verify, substitute each solution back into the original equation:

• For \( x = 0 \): \( \frac{0^{3} - 4(0)}{0^{2} + 1} = 0 \), which holds true.
• For \( x = 2 \): \( \frac{2^{3} - 4(2)}{2^{2} + 1} = \frac{8 - 8}{4 + 1} = 0 \), which is correct.
• For \( x = -2 \): \( \frac{(-2)^{3} - 4(-2)}{(-2)^{2} + 1} = \frac{-8 + 8}{4 + 1} = 0 \), also correct.

Each solution, when substituted back, results in zero, confirming their validity. Verification ensures accuracy and helps avoid mistakes in solutions.

Numerator and Denominator in Fractions

Understanding the roles of the numerator and denominator layers the foundation for solving fractional equations. In any fraction \( \frac{a}{b} \), \( a \) is the numerator and \( b \) is the denominator. A fraction equals zero only when the numerator is zero because division by any non-zero number (the denominator) will result in zero.

In our problem \( \frac{x^{3}-4x}{x^{2}+1} = 0 \), the focus initially is on the numerator \( x^{3} - 4x \). For this fraction to be zero, it is crucial that \( x^{3} - 4x \) equals zero. The denominator \( x^{2} + 1 \) cannot affect the solution directly because it cannot be zero (as squaring \( x \) and adding one will always yield a positive number).

Hence, the entire exercise revolves around ensuring the numerator equals zero while understanding the denominator plays a supportive role in maintaining the fraction's validity—not being zero to avoid undefined expressions.