Problem 32 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ \frac{5}{x-3}+\frac{x}{x+3}=\frac{19}{x^{2}-9} $$

Step-by-Step Solution

Answer

The solutions are \(x = -1 + \sqrt{5}\) and \(x = -1 - \sqrt{5}\).

Step 1: Identify a Common Denominator

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

Step 2: Rewrite Each Term with the Common Denominator

Multiply each term by an appropriate expression so that all fractions have the common denominator \((x-3)(x+3)\):- \(\frac{5}{x-3}\) becomes \(\frac{5(x+3)}{(x-3)(x+3)}\)- \(\frac{x}{x+3}\) becomes \(\frac{x(x-3)}{(x-3)(x+3)}\)- The right side, \(\frac{19}{x^2-9}\), remains the same because it already has the common denominator.

Step 3: Combine the Fractions

With all terms over the same denominator, the equation becomes:\[ \frac{5(x+3) + x(x-3)}{(x-3)(x+3)} = \frac{19}{(x-3)(x+3)} \]Combine the numerators:- \(5(x+3) = 5x + 15\)- \(x(x-3) = x^2 - 3x\)This results in the equation: \[ \frac{x^2 + 2x + 15}{(x-3)(x+3)} = \frac{19}{(x-3)(x+3)} \]

Step 4: Eliminate the Denominator

Since the denominators are identical, you can multiply both sides of the equation by \((x-3)(x+3)\) to eliminate them:\[ x^2 + 2x + 15 = 19 \].

Step 5: Solve the Resulting Quadratic Equation

Subtract 19 from both sides to set the equation to zero:\[ x^2 + 2x + 15 - 19 = 0 \]This simplifies to:\[ x^2 + 2x - 4 = 0 \]To find the values of \(x\), use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 2\), and \(c = -4\).

Step 6: Apply the Quadratic Formula

Plug the values into the quadratic formula:\[ x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-4)}}{2 \times 1} \]Calculate the discriminant and solve:- \(2^2 - 4 \times 1 \times (-4) = 4 + 16 = 20\)- \(\sqrt{20} = 2\sqrt{5}\)- \(x = \frac{-2 \pm 2\sqrt{5}}{2}\)Thus, the solutions are:\[ x = -1 \pm \sqrt{5} \].

Step 7: Check for Extraneous Solutions

Ensure the solutions do not make any original denominators zero. Neither \(-1 + \sqrt{5}\) nor \(-1 - \sqrt{5}\) result in \(x-3 = 0\) or \(x+3 = 0\), confirming they are valid solutions.

Key Concepts

Common Denominator

A common denominator is crucial when solving equations involving fractions, especially with different denominators. It helps to rewrite each fraction so they share the same lower part (denominator), allowing the numerators to be combined simply.
In our example, the initial equation has denominators of \(x-3\), \(x+3\), and \(x^2-9\). Here, recognizing that \(x^2-9\) is actually a factored form \((x-3)(x+3)\) allows us to pinpoint the common denominator easily.

• The common denominator for all parts of the equation is \((x-3)(x+3)\).
• By using this, we ensure all terms can be added or compared effortlessly, without dividing by potentially zero values.

Having a common denominator simplifies the process considerably, especially when your fractions are working within complex equations. This approach is essential for converting and solving the subsequent expressions effectively.

Factoring

Factoring is a process of breaking down an expression into a product of simpler expressions or factors.
In quadratic equations, like the one we're tackling here, understanding and applying factoring can simplify the process of finding common denominators and lead to easier solutions.

• The expression \(x^2 - 9\) factors into \((x-3)(x+3)\), which is fundamental to rewriting the terms with a common denominator.
• This factorization helps us recognize possible roots or simplifications within the equation.

Once expressions are factored, equations are more manageable since it becomes easier to manipulate and solve them. Factoring is not only beneficial in finding common denominators but also plays a role in solving quadratic equations by other methods, like rearranging or using identities.

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations and is incredibly useful when factoring is not straightforward.
Given a quadratic equation in the form \(ax^2 + bx + c = 0\), the quadratic formula provides a direct way to calculate the solutions as follows:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the example problem, the equation \(x^2 + 2x - 4 = 0\) is resolved by:

• Setting \(a = 1\), \(b = 2\), and \(c = -4\).
• Plugging these values into the formula, we calculate the discriminant \(b^2 - 4ac = 4 + 16 = 20\).
• Next, calculating the roots: \(x = \frac{-2 \pm 2\sqrt{5}}{2}\) giving solutions \(x = -1 \pm \sqrt{5}\).

This method is beneficial as it guarantees results regardless of whether or not the quadratic is easily factorable, thus offering a reliable fallback during problem-solving.

Extraneous Solutions

Extraneous solutions are potential solutions that solve the transformed equation but do not satisfy the original equation. They can arise particularly during steps where both sides of an equation are multiplied by an expression, like a common denominator.
In our exercise:

• We need to ensure our solutions \(-1 + \sqrt{5}\) and \(-1 - \sqrt{5}\) don't cause any denominator to be zero in the original equation.
• Confirming neither makes \(x-3=0\) or \(x+3=0\) ensures these solutions are valid.

It's essential to check for extraneous solutions to confirm the validity of the results, especially when transformations involve multiplying by denominators. This step ensures the roots are applicable within the context of the initial problem.