Problem 127 - College Algebra Essentials

Question

In Exercises \(127-130,\) solve each equation by the method of your choice. $$ \frac{1}{x^{2}-3 x+2}=\frac{1}{x+2}+\frac{5}{x^{2}-4} $$

Step-by-Step Solution

Answer

The feasible solution to the given equation is around \(x \approx 0.189\) and \(x \approx 1.053\). Any values of x that make any denominator in the original equation equal to zero should not be considered solutions.

Step 1: Find a Common Denominator

The least common denominator (LCD) of the fractions is found by factorizing each of the denominators and then finding the product of the highest powers of all factors obtained. The denominators \(x^{2}-3x+2\), \(x+2\), and \(x^{2}-4\) factorize to \((x-2)(x-1)\), \(x+2\), and \((x-2)(x+2)\) respectively. Hence, the LCD is \( (x-1)(x+2)(x-2)\).

Step 2: Simplify the Equation

Multiply each fraction on both sides of the equation by the LCD \( (x-1)(x+2)(x-2)\). The fractions will cancel out and leave an equation without fractions. Doing this, the equation becomes \((x+2)(x-2) + 5(x-1)(x-2) = (x-1)(x+2)\).

Step 3: Solve the Simplified Equation

Expand the left and right side of the equation to get a quadratic equation: \(x^{2} - 4 + 5x^{2} - 10x -5 = x{^2} - x - 2\). Group all terms on one side of the equation to have the equation in standard form: \(5x^{2}-11x -1 = 0\). Now use the quadratic formula, \(x=(-b±\sqrt{b^{2}-4ac})/2a\), to find the values of x.

Step 4: Check for Extraneous Solutions

Substitute the x-values obtained from the quadratic formula back into the original equation to verify that they are valid solutions. Discard any value of x that will make any denominator in the original equation equal to zero, as these are extraneous solutions.

Key Concepts

Least Common Denominator

In order to successfully solve rational equations, it is crucial to find what is known as the Least Common Denominator (LCD). This involves identifying the common denominator among the given fractions so that you can clear these fractions by multiplying through by this LCD.
By doing so, you effectively eliminate the fractions altogether, transforming the equation into an algebraic format that's easier to manage.

To find the LCD, you must:

• Factorize each denominator: Break down each denominator into its smallest factors.
• Identify all unique factors across all denominators.
• Select the highest power of each of these factors.

For example, factorizing affords you denominators like \(x^{2}-3x+2\) into \( (x-2)(x-1) \), \(x^{2}-4\) into \( (x-2)(x+2) \), and \(x+2\).
Consequently, the least common denominator for the rational equation becomes \( (x-1)(x+2)(x-2) \).
With the LCD identified, the next step involves using it to simplify the equation, allowing for smoother solving of the remaining algebraic equation.

Quadratic Formula

One of the powerful tools to solve equations, especially after transforming a rational equation into a quadratic format, is the Quadratic Formula. Quadratic equations usually appear in the standard form \(ax^{2} + bx + c = 0\). Utilizing the quadratic formula can easily find the values of \(x\) when the equation isn't factorable through simple methods like factoring or completing the square. The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} \]
By plugging in the values of \(a\), \(b\), and \(c\) from the quadratic equation, you can solve for \(x\) efficiently. For example, if you have a quadratic in the form \(5x^{2}-11x -1 = 0\), then \(a = 5\), \(b = -11\), and \(c = -1\).
Calculating within the quadratic formula renders solutions that are potential values for \(x\).
Be aware, though, that solving with the quadratic formula can sometimes yield results that may not satisfy the original equation, termed as extraneous solutions.

Extraneous Solutions

Extraneous solutions are those which arise from the solving process but do not actually satisfy the original equation. When dealing with rational equations, these solutions can often appear when potential values of \(x\) turn the denominator of a rational component into zero, thereby rendering the solution invalid.
This is why it's essential to verify each solution obtained. After solving your equation through the quadratic formula or any other method, substitute these solutions back into the original equation.

Watch out for:

• Denominators that equal zero at your found \(x\)-values.
• Undefined terms that may occur in the context of a rational equation.

By discarding any \(x\) that makes a denominator zero, you ensure that you only hold onto valid solutions. This process of validation helps reinforce a solid understanding of correctly handling rational equations and boosts accuracy in solving them.