Problem 44 - Beginning and Intermediate Algebra

Question

Use the quadratic formula to solve each equation. (All solutions for these equations are non real complex numbers.) $$ (x-1)(9 x-3)=-2 $$

Step-by-Step Solution

Answer

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

Step 1: Expand the equation

First, expand the given equation \( (x-1)(9x-3) = -2 \). Use the distributive property to expand: \( x(9x-3) - 1(9x-3) = 9x^2 - 3x - 9x + 3 \). Combine like terms to get: \( 9x^2 - 12x + 3 \).

Step 2: Set the equation to zero

Move the constant term on the left-hand side and set the equation to zero: \( 9x^2 - 12x + 3 = -2 \) Add 2 to both sides: \( 9x^2 - 12x + 5 = 0 \).

Step 3: Identify coefficients

Identify the coefficients from the quadratic equation \(9x^2 - 12x + 5 = 0\): \(a = 9, \, b = -12, \, c = 5 \).

Step 4: Apply the quadratic formula

Use the quadratic formula to solve for \(x\): \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug in the values of the coefficients: \( x = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(9)(5)}}{2(9)} \) Simplify inside the square root: \( x = \frac{12 \pm \sqrt{144 - 180}}{18} \).

Step 5: Simplify further

The equation inside the square root is: \( 144 - 180 = -36 \). Therefore, the equation simplifies to: \( x = \frac{12 \pm \sqrt{-36}}{18} \). Express \( \sqrt{-36} \) as a complex number: \( \sqrt{-36} = 6i \).

Step 6: Calculate solutions

Substitute \( 6i \) into the equation: \( x = \frac{12 \pm 6i}{18} \). Simplify the fractions: \( x = \frac{12}{18} \pm \frac{6i}{18} \). Therefore, \( x = \frac{2}{3} \pm \frac{i}{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool in algebra used to find the solutions of a quadratic equation. A quadratic equation is generally expressed in the form:
\[ ax^2 + bx + c = 0 \].
The quadratic formula itself is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].

• \(a\), \(b\), and \(c\) are the coefficients of the equation.
• \(b^2 - 4ac\) is called the discriminant.

The discriminant tells us the nature of the roots of the quadratic equation:

• If \(b^2 - 4ac > 0\), the equation has two real and distinct solutions.
• If \(b^2 - 4ac = 0\), the equation has exactly one real solution.
• If \(b^2 - 4ac < 0\), the equation has two complex solutions.

In this exercise, the discriminant is negative (<30>\(144 - 180 = -36\)), thus leading to complex number solutions.

Complex Numbers

When the discriminant (
\(b^2 - 4ac\)) is negative, we encounter square roots of negative numbers. In such cases, the solutions to the quadratic equation are not real numbers, but complex numbers.
A complex number is of the form:
\(a + bi\), where \(i\) is the imaginary unit. The imaginary unit \(i\) is defined as:
\[i = \sqrt{-1}\]
For
xample,
\(\sqrt{-36}\ = 6i\). The inclusion of \(i\) converts the solution to a combination of a real part and an imaginary part.
The solutions to the quadratic equation in the exercise are:
\(\frac{2}{3} + \frac{i}{3}\) and
\(\frac{2}{3} - \frac{i}{3}\). These are complex conjugates, meaning they have the same real part (\(\frac{2}{3}\)), but their imaginary parts are equal in magnitude and opposite in sign (\(\frac{i}{3} \) and
\( -\frac{i}{3}\)).

Solving Quadratic Equations

Let's recap the steps to solve a quadratic equation using the quadratic formula:
1. Start by expressing the quadratic equation in the standard form:
\(ax^2 + bx + c = 0\).
2. Identify the coefficients \(a\), \(b\), and \(c\).
3. Plug these coefficients into the quadratic formula:
\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
4. Compute the discriminant:
\(b^2 - 4ac\).

• If the discriminant is positive, solve for the two distinct real roots.
• If the discriminant is zero, solve for the single real root.
• If the discriminant is negative (as in our exercise), solve for the complex roots.

For our specific exercise, after calculating, we have:
\( x = \frac{12 \pm 6i}{18} \). By simplifying further:
\( x = \frac{2}{3} \pm \frac{i}{3} \). These are the final solutions in a complex form.