Problem 41 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ x(3 x+4)=-2 $$

Step-by-Step Solution

Answer

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

Step 1: Expand and write in standard form

First, expand the left-hand side and move all terms to one side to set the equation to zero.\begin{align*} x(3x + 4) &= -2 \ 3x^2 + 4x + 2 &= 0 \rightdalign*This is now in the form of a quadratic equation: \( ax^2 + bx + c = 0 \).

Step 2: Identify coefficients

Identify the coefficients \( a \), \( b \), and \( c \) from the standard form equation \( 3x^2 + 4x + 2 = 0 \).\( a = 3 \) \( b = 4 \) \( c = 2 \)

Step 3: Write the quadratic formula

Recall the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Step 4: Calculate the discriminant

Calculate the discriminant \( b^2 - 4ac \).\begin{align*}b^2 - 4ac &= 4^2 - 4(3)(2) \ &= 16 - 24 \ &= -8 \rightdalign*Since the discriminant is negative, the solutions will be non-real (complex) numbers.

Step 5: Substitute into the quadratic formula

Substitute \( a \), \( b \), and the discriminant into the quadratic formula and simplify.\begin{align*}x &= \frac{-4 \pm \sqrt{-8}}{2(3)} \ &= \frac{-4 \pm \sqrt{8i^2}}{6} \ &= \frac{-4 \pm 2\sqrt{2} i}{6} \ &= \frac{-2 \pm \sqrt{2} i}{3} \rightdalign*

Step 6: Write the final solutions

The solutions to the quadratic equation are: \( x = \frac{-2 + \sqrt{2} i}{3} \) and \( x = \frac{-2 - \sqrt{2} i}{3} \)

Key Concepts

complex numbers

Complex numbers are an extension of the real numbers and include the imaginary unit, denoted as \(i\). The imaginary unit satisfies the equation \(i^2 = -1\), which means the square of \(i\) is -1. Complex numbers are in the form \(a + bi\), where \(a\) and \(b\) are real numbers.
The number \(a\) is called the real part, and \(b\) is the imaginary part.
When solving quadratic equations, if the discriminant (\(b^2 - 4ac\)) is negative, the solutions will be complex because the square root of a negative number involves \(i\).
For example, in the equation \( x(3x + 4) = -2 \), after expanding and setting it to standard form, the discriminant (\(b^2 - 4ac\)) was calculated to be -8.
This negative result (\(-8\)) indicates that the solutions will be non-real and complex numbers.

discriminant

The discriminant is a key component in the quadratic formula and helps to determine the nature of the roots of a quadratic equation. It is given by the expression \(b^2 - 4ac\).
The value of the discriminant reveals the following:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, the quadratic equation has exactly one real solution (a repeated root).
• If the discriminant is negative, the quadratic equation has two complex solutions.

In the provided equation \( x(3x + 4) = -2 \), after expanding it and rewriting it in standard form, it became \(3x^2 + 4x + 2 = 0\).
The discriminant is calculated as follows:

• \(b = 4\)
• \(a = 3\)
• \(c = 2\)
• \(b^2 - 4ac = 4^2 - 4(3)(2) = 16 - 24 = -8\)

Since the discriminant is -8 (a negative number), the quadratic equation has two complex solutions.

quadratic equation

A quadratic equation is a polynomial equation of degree 2, generally in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a ≠ 0\).
The solutions to a quadratic equation can be found using various methods, such as factoring, completing the square, or using the quadratic formula.
The quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), allows us to find the solutions efficiently by plugging in the values of \(a\), \(b\), and \(c\).
For the given equation \( x(3x + 4) = -2 \):

• Step 1: Expand and write in standard form: \(3x^2 + 4x + 2 = 0\)
• Step 2: Identify coefficients \(a = 3\), \(b = 4\), \(c = 2\)
• Step 3: Apply the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
• Step 4: Calculate discriminant \(b^2 - 4ac = -8\)
• Step 5: Substitute into the formula: \(x = \frac{-4 \pm \sqrt{-8}}{6} = \frac{-2 \pm \sqrt{2}i}{3}\)
• Step 6: Write the final solutions: \(x = \frac{-2 + \sqrt{2} i}{3}\) and \(x = \frac{-2 - \sqrt{2} i}{3}\)

Because the discriminant is negative, the solutions are non-real complex numbers.