Problem 13 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions of each equation. Do not use a calculator. $$7 x^{3}+x=0$$

Step-by-Step Solution

Answer

Solutions are \(x = 0\), \(x = \frac{i\sqrt{7}}{7}\), and \(x = -\frac{i\sqrt{7}}{7}\).

Step 1: Factor the Equation

The given equation is \(7x^3 + x = 0\). Notice that \(x\) is a common factor in both terms. Factor \(x\) out of the equation:\[x(7x^2 + 1) = 0\].

Step 2: Solve for the First Solution

According to the Zero Product Property, if a product of factors equals zero, then at least one of the factors must be zero. Set the first factor equal to zero: \[x = 0\]. This is one of the solutions.

Step 3: Solve the Quadratic Equation

Now solve the second factor: \[7x^2 + 1 = 0\].Subtract 1 from both sides: \[7x^2 = -1\]. Divide both sides by 7 to isolate \(x^2\): \[x^2 = -\frac{1}{7}\].

Step 4: Find the Complex Solutions

To find \(x\), take the square root of both sides:\[x = \pm \sqrt{-\frac{1}{7}}\]. The square root of a negative number introduces the imaginary unit \(i\), defined as \(i = \sqrt{-1}\). Thus,\[x = \pm \frac{i}{\sqrt{7}}\]. Rationalize the denominator to get the final complex numbers:\[x = \pm \frac{i\sqrt{7}}{7}\].

Key Concepts

Zero Product Property

The Zero Product Property is an essential tool in solving polynomial equations. It states that if the product of multiple factors is zero, then at least one of the factors must be zero. This property is what allows us to break down equations into simpler parts and solve for their roots or solutions.

Let's consider the equation from the exercise:

• Original equation: \( 7x^3 + x = 0 \)
• Factored form: \( x(7x^2 + 1) = 0 \)

Using the Zero Product Property, we set each factor equal to zero independently:

• \( x = 0 \)
• \( 7x^2 + 1 = 0 \)

By setting the factors to zero individually, we enable the detection of all possible solutions. This principle simplifies the process of solving polynomial equations significantly.

Polynomial Equations

Polynomial equations comprise terms with variables raised to whole number exponents. They can range from simple linear equations to more complex cubic ones like the one in the exercise. Understanding polynomials is important for solving a variety of math problems.

In our exercise, the polynomial equation is \( 7x^3 + x = 0 \). This is a cubic equation due to the presence of the term \( 7x^3 \).

To solve polynomial equations, one often needs to factor the polynomial, as shown in the exercise.

• The first term is \( x \), a factor common to both terms, so it is factored out.
• This leads to a simpler product, \( x(7x^2 + 1) = 0 \), making it easier to identify possible zeroes.

Recognizing patterns in polynomial equations, like common factors, helps in factoring, thereby simplifying and solving them more efficiently.

Imaginary Unit

The imaginary unit, denoted as \( i \), is a pivotal concept in complex numbers. Defined as \( i = \sqrt{-1} \), it is used to express the square root of negative numbers, which don't have real solutions. This concept is crucial when solving quadratic equations with negative discriminants.

In our exercise, the equation \( 7x^2 + 1 = 0 \) turns into \( x^2 = -\frac{1}{7} \). To solve this, we take the square root of both sides:

• \( x = \pm \sqrt{-\frac{1}{7}} \)
• Using the imaginary unit, \( x = \pm \frac{i}{\sqrt{7}} \)

These new solutions involve imaginary numbers, expanding the set of possible solutions beyond real numbers. Understanding the imaginary unit opens the door to complex numbers and a broader class of solutions.

Quadratic Equations

Quadratic equations are polynomials of the second degree, usually in the form \( ax^2 + bx + c = 0 \). Understanding and solving these equations is a common step in mathematics. They often involve different processes, such as factoring, using the quadratic formula, or completing the square.

In our exercise, the equation derived from factoring is \( 7x^2 + 1 = 0 \). Normally, one would move to use further techniques like:

• Isolating the quadratic term \( x^2 \) by subtracting and dividing:
• \( x^2 = -\frac{1}{7} \)
• Finding the roots by taking square roots, where the imaginary unit becomes relevant.

Quadratic equations can be resolved using various strategies, reflecting their importance in finding solutions that may be real or complex. By comprehending quadratic equations thoroughly, students are prepared to tackle diverse mathematical problems.