Problem 26 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\frac{3 x^{2}}{x-1}+2=\frac{x}{x-1}$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{2}{3}\) and \(x = -1\).

Step 1: Identify the Common Denominator

Both sides of the equation have a common denominator of \(x-1\). Rewrite the equation to have the same denominator for easier manipulation: \[ \frac{3 x^{2}}{x-1}+2=\frac{x}{x-1} \]

Step 2: Multiply Both Sides by the Common Denominator

Eliminate the denominator by multiplying every term by \(x-1\): \[ (x-1) \times \frac{3 x^{2}}{x-1} + 2 \times (x-1) = (x-1) \times \frac{x}{x-1} \] This simplifies to: \[ 3x^2 + 2(x-1) = x \]

Step 3: Distribute and Combine Like Terms

Distribute the 2 on the left side and simplify: \[ 3x^2 + 2x - 2 = x \] Now combine like terms: \[ 3x^2 + 2x - 2 - x = 0 \] This simplifies to: \[ 3x^2 + x - 2 = 0 \]

Step 4: Solve the Quadratic Equation

Solve the quadratic equation \(3x^2 + x - 2 = 0\) using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = 1\), and \(c = -2\): \[ x = \frac{-1 \pm \sqrt{1^2 - 4 \times 3 \times (-2)}}{2 \times 3} \] Simplify under the square root: \[ x = \frac{-1 \pm \sqrt{1 + 24}}{6} \] This simplifies further to \[ x = \frac{-1 \pm \sqrt{25}}{6} \] So: \[ x = \frac{-1 \pm 5}{6} \]

Step 5: Determine the Roots

Calculate the two possible solutions: \[ x = \frac{-1 + 5}{6} = \frac{4}{6} = \frac{2}{3} \] and \[ x = \frac{-1 - 5}{6} = \frac{-6}{6} = -1 \]

Step 6: Check for Extraneous Solutions

Verify that neither solution makes the original equation undefined. Since the denominator \(x - 1\) is zero when \(x = 1\), check if either solution is 1. Neither \(x = \frac{2}{3}\) nor \(x = -1\) make the denominator zero. Hence, both solutions are valid.

Key Concepts

Common Denominator

To solve equations involving fractions, finding a common denominator is crucial. It allows us to simplify and manipulate the equation more easily. In the given exercise, both sides of the equation share a denominator of \(x-1\). By rewriting the equation with the same denominator, we can cancel out the denominators and work with simpler expressions.
Remember, this step is essential because it transforms a complex fraction equation into a more manageable polynomial equation. Practically, it means:

• Identifying the common denominator.
• Multiplying through by this common denominator to eliminate it.

This process helps in isolating the variable and solving the equation more straightforwardly.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\). In the exercise, after eliminating the common denominator, we obtain a simplified quadratic equation, \(3x^2 + x - 2 = 0\). Quadratic equations can have real or complex solutions depending on their discriminant:
\[discriminant = b^2 - 4ac\]
If the discriminant is:

• Positive, there are two distinct real solutions.
• Zero, there is exactly one real solution.
• Negative, there are no real solutions (only complex solutions).

This nature of quadratic equations makes it necessary to understand the shape and position of their graphs, which are parabolas.

Quadratic Formula

The quadratic formula is a universal tool for solving quadratic equations. It is derived from the process of completing the square. The formula is given as:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
In the exercise, it helps us solve the equation \(3x^2 + x - 2 = 0\). Using values \(a = 3\), \(b = 1\), and \(c = -2\), we plug these into the formula:

• Calculate the discriminant: \(\sqrt{1 + 24} = \sqrt{25}\)
• Apply \(\pm\) operation: \(\frac{-1 \pm 5}{6}\)
• Find solutions: \(x = \frac{4}{6} = \frac{2}{3}\) and \(x = \frac{-6}{6} = -1\)

The quadratic formula ensures we find all possible solutions for the quadratic equation.

Extraneous Solutions

Extraneous solutions are solutions that emerge during the process of solving an equation but do not satisfy the original equation. They are common in equations involving fractions or logarithms. To check for extraneous solutions:
1. Solve the equation as usual.
2. Substitute the solutions back into the original equation.
In the exercise, we derived solutions \(x = \frac{2}{3}\) and \(x = -1\). We then checked to ensure neither makes the denominator zero, verifying they are valid. This step is crucial because sometimes introduced operations (like squaring or multiplying by a variable) can create invalid solutions. Always verify your solutions to ensure they satisfy the initial equation.