Problem 27 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation by hand. $$\frac{x}{x-3}+\frac{4}{x+3}=\frac{18}{x^{2}-9}$$

Step-by-Step Solution

Answer

The solution is \(x = -10\).

Step 1: Identify Common Denominator

The denominators are \(x-3\), \(x+3\), and \(x^2-9\). Notice that \(x^2-9\) can be factored as \((x-3)(x+3)\). Therefore, \((x-3)(x+3)\) is the common denominator for all terms in the equation.

Step 2: Rewrite the Equation with Common Denominator

Rewrite each term with the common denominator \((x-3)(x+3)\):- \(\frac{x}{x-3} = \frac{x(x+3)}{(x-3)(x+3)}\)- \(\frac{4}{x+3} = \frac{4(x-3)}{(x+3)(x-3)}\)- \(\frac{18}{x^2-9} = \frac{18}{(x-3)(x+3)}\)This gives \(\frac{x(x+3) + 4(x-3)}{(x-3)(x+3)} = \frac{18}{(x-3)(x+3)}\).

Step 3: Eliminate the Denominator

Since the denominators on both sides of the equation are identical, eliminate them by setting the numerators equal:\[x(x+3) + 4(x-3) = 18\]

Step 4: Expand and Simplify the Numerator

Expand both terms in the numerator:- \(x(x+3) = x^2 + 3x\)- \(4(x-3) = 4x - 12\)Combine the terms to get:\[x^2 + 3x + 4x - 12 = x^2 + 7x - 12\]

Step 5: Solve the Quadratic Equation

Set the expanded numerator equal to the constant from Step 3:\[x^2 + 7x - 12 = 18\]Simplify to:\[x^2 + 7x - 30 = 0\]

Step 6: Apply the Quadratic Formula

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 7\), and \(c = -30\):- Calculate the discriminant: \(b^2 - 4ac = 7^2 - 4(1)(-30) = 49 + 120 = 169\).Since the discriminant is a perfect square, the roots will be real.- Calculate the solutions:\[x = \frac{-7 \pm \sqrt{169}}{2} = \frac{-7 \pm 13}{2}\]This gives two solutions: \(x = 3\) and \(x = -10\).

Step 7: Check for Extraneous Solutions

Since the original equation had fractions with denominators \(x-3\) and \(x+3\), solutions that make these denominators zero are extraneous. Thus, \(x = 3\) is not a valid solution due to division by zero. Therefore, the only valid solution is \(x = -10\).

Key Concepts

Quadratic Equation

Quadratic equations pop up often in math problems, including when finding complex number solutions. A quadratic equation is typically solved when it is in the standard form: \[ ax^2 + bx + c = 0 \] In this form, \(a\), \(b\), and \(c\) are known numbers, with \(a eq 0\). The solutions or "roots" of this equation can be found using various methods like factorization, completing the square, and the quadratic formula. Each approach has its own use depending on the specifics of the quadratic equation you are dealing with.
A handy tool is the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] It provides the roots of the quadratic equation. The term inside the square root, \(b^2 - 4ac\), is called the discriminant.
**Understanding the Discriminant** - If the discriminant is positive, the equation has two distinct real roots. - If it’s zero, there’s exactly one real root (a repeated root). - If it’s negative, the roots will be complex conjugates. This particular problem's discriminant was 169, indicating two distinct real solutions. By substituting values of \(b\), \(a\), and \(c\) into the formula, we solve the problem efficiently even for complex numbers.

Common Denominator

Adding or subtracting fractions requires a common denominator. This is key when working with rational expressions, like in our original equation with complex numbers.
In simple terms, a common denominator is just a shared multiple of the denominators you are working with. The purpose is to rewrite fractions so they have a uniform bottom part, simplifying the addition or subtraction process.
**Finding a Common Denominator** - Identify the denominators you have: In the given problem, these were \(x-3\), \(x+3\), and \(x^2 - 9\). - Factor these into simpler parts where possible. Here, \(x^2 - 9\) was factored into \((x-3)(x+3)\). This made it clear that \((x-3)(x+3)\) was our common denominator. - Rewrite each fraction using the common denominator: Multiply the numerator and denominator to match the common denominator without changing the original fraction's value. This step allows you to combine or subtract the rational expressions easily without worrying about mismatched denominators.

Extraneous Solutions

Extraneous solutions in an equation are like uninvited guests at a party. These are solutions that appear during the process of solving an equation but are not valid for the original equation.
Extraneous solutions frequently arise when we manipulate equations, such as when we square both sides or when dealing with rational expressions. Manipulating the equation might introduce solutions that do not satisfy the original conditions.
**Checking for Extraneous Solutions** - Always plug the solutions back into the original equation. This ensures they do not cause division by zero or other undefined operations. - In our problem, setting \(x = 3\) resulted in division by zero in the original denominators \(x-3\) and \((x+3)\). Hence, \(x = 3\) was deemed extraneous. - The remaining solution \(x = -10\) did not cause any problems with the denominators and was thus the valid solution. Checking for these extraneous roots is crucial. It helps ascertain which roots actually solve the original equation under all conditions provided.