Problem 28 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation by hand. $$\frac{2 x}{x-3}+\frac{4}{x+3}=\frac{24}{9-x^{2}}$$

Step-by-Step Solution

Answer

The complex solutions are \( x = \frac{-5 + \sqrt{145}}{2} \) and \( x = \frac{-5 - \sqrt{145}}{2} \).

Step 1: Rewrite the Equation

The given equation is \( \frac{2x}{x-3} + \frac{4}{x+3} = \frac{24}{9-x^2} \). Notice that \(9 - x^2\) can be factored as \((3-x)(3+x)\). Rewriting the equation gives: \[ \frac{2x}{x-3} + \frac{4}{x+3} = \frac{24}{(3-x)(3+x)}. \]

Step 2: Find a Common Denominator

The denominators are \((x-3)\), \((x+3)\), and \((3-x)(3+x)\). The least common denominator (LCD) is \((x-3)(x+3)\). Rewriting in terms of the LCD: \[ \frac{2x(x+3)}{(x-3)(x+3)} + \frac{4(x-3)}{(x-3)(x+3)} = \frac{24}{(x-3)(x+3)}. \]

Step 3: Combine the Fractions

Combine the fractions on the left side over the common denominator: \[ \frac{2x(x+3) + 4(x-3)}{(x-3)(x+3)} = \frac{24}{(x-3)(x+3)}. \]

Step 4: Equate Numerators

Since the denominators are equal, equate the numerators: \( 2x(x+3) + 4(x-3) = 24 \). Expand and simplify the equation: \( 2x^2 + 6x + 4x - 12 = 24 \). Simplify to get: \( 2x^2 + 10x - 36 = 24 \).

Step 5: Solve the Quadratic Equation

Subtract 24 from both sides to form a standard quadratic equation: \( 2x^2 + 10x - 60 = 0 \). Divide by 2: \( x^2 + 5x - 30 = 0 \). Solve using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = 5 \), \( c = -30 \).

Step 6: Calculate Discriminant

Calculate the discriminant: \( b^2 - 4ac = 5^2 - 4(1)(-30) = 25 + 120 = 145 \). The roots are \( x = \frac{-5 \pm \sqrt{145}}{2} \).

Step 7: Simplify Roots

Calculate the exact values: \( x_1 = \frac{-5 + \sqrt{145}}{2} \) and \( x_2 = \frac{-5 - \sqrt{145}}{2} \).

Step 8: Validate the Solution

Ensure that the solutions obtained do not make the original denominators zero. Substitute back into the initial terms to verify which solutions are valid.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree two in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). This formula represents a parabola when plotted on a graph, opening upwards if \( a > 0 \) and downwards if \( a < 0 \). Solving a quadratic equation involves finding the values of \( x \) that make the equation true. These are known as the roots of the equation.

To solve a quadratic equation, several methods can be used, including factoring, completing the square, and using the quadratic formula. The quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) is often used because it provides a direct way to find the roots, even when they are not rational. In the exercise provided, we ultimately used the quadratic formula to find the complex solutions for the equation.

Discriminant

The discriminant is a component of the quadratic formula located under the square root sign \( (\sqrt{b^2 - 4ac}) \). It helps determine the nature of the roots of a quadratic equation:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root (a repeated root).
• If the discriminant is negative, there are two complex conjugate roots.

In our exercise, we calculated the discriminant as \( b^2 - 4ac = 145 \), which is positive. This confirmed that the quadratic equation had two distinct real roots. Although the exercise focused on complex solutions, noting the value of the discriminant helped confirm the method of solving.

Factoring Expressions

Factoring is the process of breaking down an expression into simpler "factors" that, multiplied together, yield the original expression. In quadratic equations, factoring involves finding two binomials that produce the original quadratic when multiplied together. Not all quadratics can be factored easily, especially when the result is not a perfect square or simple integers.

In the exercise's quadratic equation \( x^2 + 5x - 30 = 0 \), direct factoring was not practical since the expression doesn't easily break down into integer factors. Instead, after manipulating the initial equation, we chose the more suitable quadratic formula approach to solve for the roots. Understanding when and how to factor can be very useful for simplifying expressions and equations, making complex problems more manageable.

Common Denominator

A common denominator is essential when combining fractions, as it allows each fraction to be expressed with a shared denominator, making them easier to add or subtract. To find the least common denominator (LCD), it is necessary to identify a term that each of the original denominators can divide into without a remainder.

In the given exercise, the denominators were \((x-3)\), \((x+3)\), and \((3-x)(3+x)\). The LCD, \((x-3)(x+3)\), allowed us to rewrite all terms to have the same denominator, making it possible to combine the fractions effectively. This step was crucial in simplifying the equation so that it could be solved using other algebraic methods. Recognizing and using a common denominator is a foundational skill in understanding and solving algebraic expressions and equations.