Problem 60 - College Algebra with Modeling and Visualization

Question

Solve. Write answers in standard form. $$ x(3 x+1)=-1 $$

Step-by-Step Solution

Answer

The solutions are \(x = -\frac{1}{6} \pm \frac{i\sqrt{11}}{6}\).

Step 1: Expand the Equation

Start by expanding the equation on the left-hand side. You have the equation \(x(3x + 1) = -1\). First, distribute \(x\) to both terms inside the parentheses:\[3x^2 + x = -1\]

Step 2: Set Equation to Standard Form

Next, move all terms to one side of the equation to set it to the standard quadratic form \(ax^2 + bx + c = 0\):\[3x^2 + x + 1 = 0\]

Step 3: Identify Coefficients

Identify the coefficients \(a\), \(b\), and \(c\) in the quadratic equation \(3x^2 + x + 1 = 0\). Here, \(a = 3\), \(b = 1\), and \(c = 1\).

Step 4: Use the Quadratic Formula

Apply the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(x\). Substitute the identified coefficients into the formula:\[x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3}\]

Step 5: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\):\[1^2 - 4 \cdot 3 \cdot 1 = 1 - 12 = -11\]Since the discriminant is negative, there are no real solutions.

Step 6: Find Complex Solutions

Since the discriminant is negative, the solutions are complex. Compute the complex solutions using the imaginary unit \(i\):\[x = \frac{-1 \pm \sqrt{-11}}{6}\]This simplifies to:\[x = \frac{-1 \pm i\sqrt{11}}{6}\]

Step 7: Write Final Answer in Standard Form

Write the complex solutions in standard form \(a + bi\) where \(i\) is the imaginary unit:\[x = -\frac{1}{6} \pm \frac{i\sqrt{11}}{6}\]

Key Concepts

Standard Form

In mathematics, especially when dealing with quadratic equations, it's crucial to format them in what is known as the "standard form." This is written as: \[ ax^2 + bx + c = 0 \] where:

• \( a \) represents the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

This format is essential for simplifying the process of solving, analyzing, and graphing quadratic equations. For our exercise, starting with \( x(3x + 1) = -1 \), expanding and reorganizing the equation helps us to write it in standard form as \( 3x^2 + x + 1 = 0 \). Having this in standard form allows us to apply solution techniques like factoring (if possible) or using the quadratic formula to find the roots of the equation.
Moving everything to one side of the equation is usually the first step in solving any algebraic equation because it sets the stage for finding the unknown values of \( x \), making the problem more straightforward.

Quadratic Formula

The quadratic formula is a powerful tool for finding solutions of quadratic equations, especially when factoring is not straightforward or possible. When you are given an equation in the standard quadratic form, you can always find its roots using:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]To use this formula, it's important to identify the values of \( a \), \( b \), and \( c \) from your standard form equation. For instance, in \( 3x^2 + x + 1 = 0 \), \( a = 3 \), \( b = 1 \), and \( c = 1 \).
Plugging these values into the formula, we compute the discriminant \( b^2 - 4ac \) first, which helps determine the nature of the roots. In our exercise, the discriminant \( 1^2 - 4 \times 3 \times 1 = -11 \) is negative, indicating complex solutions. The quadratic formula is extremely versatile, applicable to all quadratic equations, ensuring we can always find a solution whether those solutions are real or complex.

Complex Solutions

When the discriminant \( b^2 - 4ac \) of a quadratic equation is negative, it means the roots of the equation are not real but complex. Complex numbers include a real part and an imaginary part. The imaginary unit \( i \) is defined as \( \sqrt{-1} \).
In our particular solution, \[ x = \frac{-1 \pm \sqrt{-11}}{6} \]the presence of \( \sqrt{-11} \) necessitates the introduction of \( i \), which transforms the solution to:\[x = \frac{-1}{6} \pm \frac{i\sqrt{11}}{6}\]
This expresses each root in standard complex form \( a + bi \), where \( a \) is the real component and \( bi \) is the imaginary component. Complex solutions came about due to the negative discriminant, highlighting that not every quadratic equation will intersect the x-axis at real points.
Understanding complex solutions widens the scope of solving quadratics, as you realize that solutions exist not just within real numbers but also in the complex plane.