Problem 20 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\frac{7}{x^{2}}+\frac{19}{x}=6$$

Step-by-Step Solution

Answer

The solutions are \( x = 3.5 \) and \( x = -\frac{1}{3} \).

Step 1: Identify the common denominator

To solve the equation \ \ \ \ \( \frac{7}{x^{2}}+\frac{19}{x}=6 \), identify a common denominator. Here, the common denominator is \( x^2 \).

Step 2: Multiply through by the common denominator

Multiply each term by \( x^2 \) to eliminate the fractions: \ \ \ \ \( x^2 \cdot \frac{7}{x^{2}} + x^2 \cdot \frac{19}{x} = 6 \cdot x^2 \). This simplifies to: \ \ \ \ \( 7 + 19x = 6x^2 \).

Step 3: Rearrange to form a quadratic equation

Rearrange the equation to the standard quadratic form \( ax^2 + bx + c = 0 \): \ \ \ \ \( 6x^2 - 19x - 7 = 0 \).

Step 4: Use the quadratic formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 6 \), \( b = -19 \), and \( c = -7 \). Plug these into the formula: \ \ \ \ \( x = \frac{-(-19) \pm \sqrt{(-19)^2 - 4 \cdot 6 \cdot (-7)}}{2 \cdot 6} \).

Step 5: Simplify under the square root

Simplify the term under the square root: \ \ \ \ \( \sqrt{361 + 168} = \sqrt{529} = 23 \).

Step 6: Solve for x

Substitute back and simplify: \ \ \ \ \( x = \frac{19 \pm 23}{12} \). This gives two solutions: \ \ \ \ \( x = \frac{19 + 23}{12} = \frac{42}{12} = 3.5 \) and \( x = \frac{19 - 23}{12} = \frac{-4}{12} = -\frac{1}{3} \).

Step 7: Verify the solutions

Verify that both solutions satisfy the original equation. For \( x = 3.5 \): \ \ \ \ \( \frac{7}{(3.5)^2} + \frac{19}{3.5} \approx 6 \), similarly for \( x = -\frac{1}{3} \): \ \ \ \ \( \frac{7}{(-\frac{1}{3})^2} + \frac{19}{-\frac{1}{3}} \approx 6 \).

Key Concepts

Common Denominator

Understanding the concept of the common denominator is crucial when you're dealing with equations that include fractions. A common denominator allows you to combine different fractions into a single equation, simplifying the process of solving it.

In the equation \( \frac{7}{x^2} + \frac{19}{x} = 6 \), the fractions \( \frac{7}{x^2} \) and \( \frac{19}{x} \) have different denominators. To make them easier to work with, we need a common denominator. For these fractions, the common denominator is \( x^2 \).

Why \( x^2 \)? Because \( x^2 \) is the smallest expression that both \( x^2 \) and \( x \) can divide into without leaving a remainder. By multiplying through by the common denominator, we eliminate the fractions, simplifying our equation to \( 7 + 19x = 6x^2 \). This step sets us up to transform our equation into the familiar format of a quadratic equation.

Quadratic Formula

The quadratic formula is a powerful tool for solving any quadratic equation of the form \( ax^2 + bx + c = 0 \). Once you have your equation in standard form, you can find the solutions for \(x\) using: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \).

In our problem, we reformed the equation to \( 6x^2 - 19x - 7 = 0 \). Here, \( a = 6 \), \( b = -19 \), and \( c = -7 \). Plugging these values into the quadratic formula, we get: \( x = \frac{-(-19) \, \pm \, \sqrt{(-19)^2 - 4 \, \cdot \, 6 \, \cdot \, (-7)}}{2 \, \cdot \, 6} \).

Simplify under the square root: \( \ \sqrt{361 + 168} = \sqrt{529} = 23 \).

So, our solutions are: \( x = \frac{19 \pm 23}{12} \). This breaks down to: - \( x = \frac{19 + 23}{12} = 3.5 \) - \( x = \frac{19 - 23}{12} = -\frac{1}{3} \).

Equation Verification

After solving the quadratic equation, it's essential to verify that both solutions satisfy the original equation. This ensures that no mistakes were made during solving.

For \( x = 3.5 \): - Substitute \( x \) back into the original equation: \( \frac{7}{(3.5)^2} + \frac{19}{3.5} \). - Verify that this equals 6: \( \frac{7}{12.25} + \frac{19}{3.5} \approx 6 \).

For \( x = -\frac{1}{3} \): - Substitute \( x \) back into the original equation: \( \frac{7}{(-\frac{1}{3})^2} + \frac{19}{-\frac{1}{3}} \) - Verify that this equals 6: \( \frac{7}{\frac{1}{9}} + \frac{19}{-\frac{1}{3}} \ = 63 - 57 \approx 6 \).