Problem 86 - Intermediate Algebra

Question

Solve equation. If a solution is extraneous, so indicate. \(\frac{x}{x-5}+\frac{5}{x}=\frac{11}{6}\)

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = 15\), both valid.

Step 1: Eliminate Denominators

Identify the common denominator for all terms in the equation. The denominators are \(x(x-5)\), so multiply every term by \(x(x-5)\) to eliminate the fractions. The equation becomes:\[x \, \frac{x}{x-5} + x(x-5) \, \frac{5}{x} = x(x-5) \, \frac{11}{6}\]This simplifies to:\[x^2 + 5(x-5) = \frac{11}{6}x(x-5)\]

Step 2: Simplify Each Side

Simplify both sides of the equation by carrying out the multiplications:- Left side: \(x^2 + 5x - 25\)- Substitute into the equation: \[x^2 + 5x - 25 = \frac{11}{6}(x^2 - 5x)\]Multiply out the right side:\[x^2 + 5x - 25 = \frac{11}{6}x^2 - \frac{55}{6}x\]

Step 3: Equate and Clear Fractions

To clear the fraction, multiply through by 6 (the denominator of \(\frac{11}{6}\)) to eliminate the fraction on the right side:\[6(x^2 + 5x - 25) = 11x^2 - 55x\]This simplifies to:\[6x^2 + 30x - 150 = 11x^2 - 55x\]

Step 4: Rearrange to Form a Quadratic

Bring all terms to one side to form a quadratic equation:\[6x^2 + 30x - 150 - 11x^2 + 55x = 0\]Combine like terms:\[-5x^2 + 85x - 150 = 0\]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula, where \(a = -5\), \(b = 85\), and \(c = -150\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Calculate the discriminant:\[b^2 - 4ac = 85^2 - 4(-5)(-150) = 7225 - 3000 = 4225\]Since the square root of 4225 is 65, plug into the formula:\[x = \frac{-85 \pm 65}{-10}\]This gives solutions for \(x\):\[x = 2, \quad x = 15\]

Step 6: Check Solutions for Extraneous Results

Substitute both solutions back into the original equation to confirm they do not make any denominators zero:- If \(x = 2\), the denominators are 2 - 5 and 2, both non-zero.- If \(x = 15\), the denominators are 15 - 5 and 15, both non-zero.Neither solution is extraneous.

Key Concepts

Understanding Extraneous Solutions

When solving rational equations, you may come across solutions that don't actually work in the original equation. These are called extraneous solutions. They often appear when you manipulate the equation to eliminate fractions, such as multiplying through by a common denominator.

Here's why they occur:

• When you multiply both sides of an equation by a variable expression, you might inadvertently introduce values that are not valid for the original equation.
• Particularly, if the solution makes any denominator in the original equation equal to zero, it is considered extraneous.

It is essential to check all potential solutions back in the original equation to ensure they do not result in undefined operations, like division by zero. In our example, neither solution, 2 nor 15, resulted in a zero denominator; hence neither was extraneous.

Finding a Common Denominator

Working with rational equations often involves finding a common denominator to combine fractions. This is a key step in solving these equations because it allows us to eliminate fractions, which can simplify the problem.

For the given equation \(\frac{x}{x-5}+\frac{5}{x}=\frac{11}{6}\), the denominators are \(x\) and \(x-5\). The common denominator here is \(x(x-5)\), as it can make both fractions have the same denominator.

• The process involves multiplying every term by this common denominator to get rid of the fractions.
• This simplifies the equation to an easier form that's free of fractions and can be further simplified.

This step is crucial because it transforms a potentially complex rational equation into an algebraic equation that is often easier to solve.

Applying the Quadratic Formula

Once a rational equation is transformed into a quadratic equation, solving it involves another critical concept: the quadratic formula. This formula helps find the roots of any quadratic equation of the form \(ax^2 + bx + c = 0\).

The quadratic formula is written as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

• Where \(a\), \(b\), and \(c\) are the coefficients of the terms in the quadratic equation.
• The formula calculates the values of \(x\) that make the quadratic equation zero.

In the problem we worked on, \(a = -5\), \(b = 85\), and \(c = -150\), plugging these into the formula gives the solutions for \(x\). This approach is especially useful because it applies to any quadratic, regardless of whether it can be factored easily or not.

Understanding the Discriminant

The discriminant in a quadratic equation is an important component when applying the quadratic formula. It is the part under the square root: \(b^2 - 4ac\).

This value helps determine the nature and number of the solutions of a quadratic equation:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it's negative, there are no real solutions (the solutions are complex numbers).

In the given problem, the discriminant calculated to be 4225, which is positive, signifies two distinct real solutions for the equation. This is crucial as it provides insight into what kind of solutions to expect, even before calculating them with the quadratic formula.