Problem 55 - Precalculus: Functions and Graphs

Question

Find the solutions of the equation. $$x^{3}+3 x^{2}+4 x=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = \frac{-3 + i\sqrt{7}}{2}\), and \(x = \frac{-3 - i\sqrt{7}}{2}\).

Step 1: Factor the equation

The given equation is \(x^3 + 3x^2 + 4x = 0\). Notice that each term has an \(x\) in common. Factor out \(x\) from the entire equation:\[x(x^2 + 3x + 4) = 0\]This gives us one solution directly: \(x = 0\). The other solutions will come from solving the quadratic equation \(x^2 + 3x + 4 = 0\).

Step 2: Solve the quadratic equation

Now, solve the quadratic equation \(x^2 + 3x + 4 = 0\) using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For the equation \(x^2 + 3x + 4 = 0\), \(a = 1\), \(b = 3\), and \(c = 4\). Substitute these values into the formula:\[x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}\]Simplify the expression inside the square root: \(3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7\).Thus, the solutions are:\[x = \frac{-3 \pm \sqrt{-7}}{2}\]Since the discriminant is negative, the roots are complex: \[x = \frac{-3 \pm i\sqrt{7}}{2}\]

Step 3: List all the solutions

We have found three solutions for the original cubic equation. The solutions are:1. \(x = 0\) (from factoring the original equation)2. \(x = \frac{-3 + i\sqrt{7}}{2}\) (complex root from solving the quadratic)3. \(x = \frac{-3 - i\sqrt{7}}{2}\) (complex root from solving the quadratic)

Key Concepts

Factoring Polynomials

Factoring polynomials is a method used to simplify polynomial expressions by finding factors that are common to all terms. In the polynomial equation \(x^3 + 3x^2 + 4x = 0\), notice that each term shares an \(x\). By factoring out the \(x\), the equation reduces to \(x(x^2 + 3x + 4) = 0\).

• Factoring allows us to break the polynomial into simpler parts, making it easier to find solutions.
• Once we factor out the common term, we can solve for \(x\) by setting each factor equal to zero.

Thus, the factor \(x = 0\) gives us a direct solution. The quadratic \(x^2 + 3x + 4 = 0\) must be solved separately to find other solutions.

Quadratic Formula

The quadratic formula is a powerful tool used to find solutions to quadratic equations of the form \(ax^2 + bx + c = 0\). For the quadratic \(x^2 + 3x + 4 = 0\), we apply the quadratic formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
For our quadratic, \(a = 1\), \(b = 3\), and \(c = 4\), which we plug into the formula:

• \(b^2 - 4ac\) is the discriminant. In this case, it is \(3^2 - 4 \cdot 1 \cdot 4 = 9 - 16 = -7\).
• A negative discriminant results in complex roots, meaning the solutions are not real numbers.

This yields the solutions \(x = \frac{-3 \pm \sqrt{-7}}{2}\).

Complex Numbers

Complex numbers are numbers that have both a real and an imaginary component, expressed in the form \(a + bi\), where \(i\) is the square root of \(-1\).

• In our solutions such as \(x = \frac{-3 \pm i\sqrt{7}}{2}\), the \(i\sqrt{7}\) represents the imaginary part.
• These come into play when dealing with negative discriminants in the quadratic formula, as we saw in \(x^2 + 3x + 4 = 0\).

Understanding how to work with complex numbers is essential when faced with quadratic equations that cannot be solved by traditional real number methods.

Solutions of Equations

Finding solutions to equations means determining the values of variables that make the equation true. In the case of\(x^3 + 3x^2 + 4x = 0\), solutions encompass both real and complex roots.

• The factored \(x\) gives a clear real solution: \(x = 0\).
• The complex solutions \(x = \frac{-3 + i\sqrt{7}}{2}\) and \(x = \frac{-3 - i\sqrt{7}}{2}\) arise from the quadratic formula result of \(x^2 + 3x + 4 = 0\).

Root solutions can be verified by plugging them back into the original equation, ensuring the left-hand side equals the right-hand side, confirming the accuracy of these solutions.