Problem 79 - Beginning and Intermediate Algebra

Question

The following exercises are not grouped by type. Solve each equation. (Exercises 83 and 84 require knowledge of complex numbers.) \(6=7(2 w-3)^{-1}+3(2 w-3)^{-2}\)

Step-by-Step Solution

Answer

The solutions to the equation are \(w = 2.25\) and \(w = 1.25\).

Step 1: Identify the Variable

Recognize that the variable in the equation is \(w\). The equation is given in terms of \(2w-3\).

Step 2: Introduce a Substitution

Let \(u = 2w-3\). Rewrite the equation in terms of \(u\): \(6 = 7u^{-1} + 3u^{-2}\).

Step 3: Simplify the Expression

Rewrite the right-hand side of the equation: \(7u^{-1} + 3u^{-2}\) becomes \( \frac{7}{u} + \frac{3}{u^2} \).

Step 4: Multiply by \(u^2\)

To eliminate the fractions, multiply both sides of the equation by \(u^2\). This yields: \(6u^2 = 7u + 3\).

Step 5: Rearrange to Form a Quadratic

Move all terms to one side to form a standard quadratic equation: \(6u^2 - 7u - 3 = 0\).

Step 6: Solve the Quadratic

Use the quadratic formula \(u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 6\), \(b = -7\), and \(c = -3\). Calculate the discriminant: \(b^2 - 4ac = (-7)^2 - 4(6)(-3) = 49 + 72 = 121\).

Step 7: Apply the Quadratic Formula

Substitute into the quadratic formula: \(u = \frac{7 \pm \sqrt{121}}{12}\). Simplify to find \(u = \frac{7 \pm 11}{12}\). Thus, the solutions are \(u = 1.5\) or \(u = -\frac{1}{2}\).

Step 8: Back-Substitute to Find \(w\)

Recall that \(u = 2w - 3\). For \(u = 1.5\): \(1.5 = 2w - 3\) \(2w = 4.5\) \(w = 2.25\). For \(u = -\frac{1}{2}\): \(-\frac{1}{2} = 2w - 3\) \(2w = 2.5\) \(w = 1.25\).

Key Concepts

Complex Numbers

Complex numbers are numbers that include a real part and an imaginary part. They are written in the form of \(a + bi\) where \(a\) is the real part, \(b\) is the imaginary part, and \(i\) is the imaginary unit, defined as \(i = \sqrt{-1}\). When solving equations, complex numbers can appear especially when the discriminant (part of the quadratic formula) is negative. This means the equation has no real solution but instead has solutions involving \(i\). Complex numbers extend the idea of one-dimensional number lines to two-dimensional planes (complex plane). It's crucial to understand complex arithmetic, including addition, subtraction, multiplication, and division.

Quadratic Formula

The quadratic formula is a mathematical solution used to solve quadratic equations of the form \(ax^2 + bx + c = 0\). The formula is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]. This formula derives from completing the square in the equation. The discriminant, \(b^2 - 4ac\), is essential for determining the nature of the roots of the equation:

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

Using the quadratic formula, you substitute the coefficients from the quadratic equation into the formula to find its solutions.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying mathematical expressions and equations. It involves using various algebra rules and operations to solve for the unknown variable. Steps typically include:

• Combining like terms: Terms with the same variables raised to the same power are combined.
• Using the distributive property: Expanding expressions by distributing multipliers across terms within parentheses.
• Isolating the variable: Moving terms from one side of the equation to the other to isolate the variable of interest.
• Substitution: Replacing one variable or expression with another equivalent one to simplify the equation or solve it. An example is the substitution used in the provided exercise, where \(u = 2w - 3\) simplified the original equation.

Mastering algebraic manipulation is fundamental to solving more complex equations, as demonstrated in solving the quadratic equation within the provided exercise.