Problem 37 - College Algebra with Modeling and Visualization

Question

Solve the polynomial equation. $$ x^{4}=x^{3}-4 x^{2} $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \), \( x = \frac{1 + i\sqrt{15}}{2} \), and \( x = \frac{1 - i\sqrt{15}}{2} \).

Step 1: Set Equation to Zero

Start by moving all terms to one side of the equation to set it to zero. The original equation is \( x^4 = x^3 - 4x^2 \). Subtract \( x^3 \) and add \( 4x^2 \) to both sides to get: \( x^4 - x^3 + 4x^2 = 0 \).

Step 2: Factor Out the Common Term

Identify a common factor in the equation. All terms contain at least \( x^2 \). Factor \( x^2 \) from the equation: \( x^2(x^2 - x + 4) = 0 \). This gives a product of terms equal to zero.

Step 3: Solve Each Factor Separately

For the factored equation \( x^2(x^2 - x + 4) = 0 \), solve the product by zero property: \[ x^2 = 0 \] and \[ x^2 - x + 4 = 0 \].

Step 4: Solve for x from x^2 = 0

Solve \( x^2 = 0 \). When a square equals zero, the solution is \( x = 0 \).

Step 5: Solve the Quadratic x^2 - x + 4 = 0

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve the quadratic \( x^2 - x + 4 = 0 \), with \( a = 1, b = -1, c = 4 \).

Step 6: Calculate the Discriminant

First, calculate the discriminant: \( b^2 - 4ac = (-1)^2 - 4 \times 1 \times 4 = 1 - 16 = -15 \). The discriminant is negative, indicating complex roots.

Step 7: Find Complex Roots

Compute the complex roots using the quadratic formula; \( x = \frac{-(-1) \pm \sqrt{-15}}{2 \cdot 1} \) simplifies to \( x = \frac{1 \pm i\sqrt{15}}{2} \). The solutions are \( x = \frac{1 + i\sqrt{15}}{2} \) and \( x = \frac{1 - i\sqrt{15}}{2} \).

Key Concepts

Factoring Polynomials

Factoring polynomials is a crucial skill in algebra that simplifies equations and exposes potential solutions. Begin by examining the polynomial and identifying any common factors across all terms. In the example equation \(x^4 - x^3 + 4x^2 = 0\), observe that each term includes at least \(x^2\) as a factor. Extracting \(x^2\) results in a simpler form: \(x^2(x^2 - x + 4) = 0\). This separation into factors allows us to apply the zero product property, setting each factor to zero individually.

• The zero product property states that if a product of terms equals zero, at least one of the terms must also be zero.
• This principle enables us to solve \(x^2 = 0\) and \(x^2 - x + 4 = 0\) separately.

Factoring is highly useful not only in solving equations but also in simplifying expressions for further mathematical operations.

Complex Roots

Complex roots often emerge when solving polynomial equations with a negative discriminant. Such a discriminant tells us that real number solutions do not exist, leading us into the realm of complex numbers. In the equation \(x^2 - x + 4 = 0\), we find the discriminant \(b^2 - 4ac = -15\). As this value is negative, the quadratic equation possesses roots that include imaginary numbers.

• Complex numbers consist of a real part and an imaginary part, expressed as \(a + bi\), where \(i\) is the square root of \(-1\).
• Imaginary roots appear in pairs, thus ensuring polynomial equations have a balanced set of solutions.

To solve for complex roots, the quadratic formula is adjusted to accommodate the imaginary part. This results in solutions that incorporate \(i\), such as \(x = \frac{1 \pm i\sqrt{15}}{2}\).

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). Applicable to any quadratic equation, the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) provides direct access to its roots. Let's break down this tool in our provided equation:

• Identify \(a = 1\), \(b = -1\), and \(c = 4\).
• Calculate the discriminant \(b^2 - 4ac\) to predict the nature of the roots.
• A negative discriminant, as seen here, indicates the presence of complex roots.

Using these values, the quadratic formula helps us find solutions such as \(x = \frac{1 + i\sqrt{15}}{2}\) and \(x = \frac{1 - i\sqrt{15}}{2}\). By providing a structured approach, the quadratic formula simplifies the solving of quadratic equations and handles complex solutions with ease.