Problem 18 - A Graphical Approach to Precalculus with Limits

Question

Find all complex solutions for each equation by hand. $$1-\frac{3}{x}-\frac{10}{x^{2}}=0$$

Step-by-Step Solution

Answer

The complex solutions are \(x = 5\) and \(x = -2\).

Step 1: Eliminate the Fractions

To eliminate the fractions, multiply the entire equation by \(x^2\). This gives you:\[x^2 - 3x - 10 = 0\]

Step 2: Recognize the Quadratic Form

The equation \(x^2 - 3x - 10 = 0\) is now a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -3\), and \(c = -10\).

Step 3: Apply the Quadratic Formula

Use the quadratic formula to solve for \(x\):\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Plug in the values: \(a = 1\), \(b = -3\), and \(c = -10\):\[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}\]\[x = \frac{3 \pm \sqrt{9 + 40}}{2}\]

Step 4: Simplify Inside the Square Root

Calculate the value inside the square root:\[9 + 40 = 49\]Therefore, the square root is \(\sqrt{49} = 7\).

Step 5: Determine the Roots

Using the simplified square root, continue solving:\[x = \frac{3 \pm 7}{2}\]This gives us two potential solutions:\[x = \frac{3 + 7}{2} = 5\]\[x = \frac{3 - 7}{2} = -2\]

Step 6: Verify the Solutions

Substitute \(x = 5\) and \(x = -2\) back into the original equation to ensure both satisfy it:- For \(x = 5\): \[1 - \frac{3}{5} - \frac{10}{25} = 1 - \frac{3}{5} - \frac{2}{5} = 1 - 1 = 0\]- For \(x = -2\): \[1 - \frac{3}{-2} - \frac{10}{(-2)^2} = 1 + \frac{3}{2} - \frac{10}{4} = 1.5 - 2.5 = 0\]Both solutions satisfy the equation.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful mathematical tool that helps us solve quadratic equations. A quadratic equation generally takes the form \( ax^2 + bx + c = 0 \). The formula to find the roots (solutions) of this equation is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It efficiently finds complex (and real) solutions without the need for factoring. Here's how it works:

• Identify the coefficients: In the equation \( ax^2 + bx + c = 0 \), \( a \), \( b \), and \( c \) are the coefficients.
• Plug these coefficients into the quadratic formula.
• The \( \pm \) symbol indicates that there will be two solutions: one involving addition and the other subtraction.
• The expression under the square root sign \( b^2 - 4ac \) is called the discriminant. It determines the nature (real or complex) of the solutions.

When you calculate the discriminant:- If it's positive, there are two distinct real solutions.- If zero, there's exactly one real solution.- If negative, the solutions are complex, involving imaginary numbers.

Fractions Elimination

Dealing with fractions in equations can be tricky, so it's often helpful to eliminate them for simplicity. To do this, find a common denominator or multiply through by a term that cancels the denominators. In our exercise:
Multiplying the entire equation \( 1 - \frac{3}{x} - \frac{10}{x^2} = 0 \) by \( x^2 \):

• Removes the fractions, resulting in the equation \( x^2 - 3x - 10 = 0 \).

This method transforms the problem into a more manageable quadratic equation without fractions. Here's why that's helpful:

• Easier calculations: Quadratic equations are often simpler to solve than ones with fractions.
• Fewer mistakes: Working with fractions can lead to errors in arithmetic.
• Clearer expressions: Without fractions, equations are typically easier to interpret and manipulate.

Removing fractions early on can save time and reduce frustration in the problem-solving process.

Verification of Solutions

After finding the solution to an equation, it's essential to verify that the solutions actually satisfy the original equation. This verification ensures there were no mistakes in the process. Here's how to verify solutions:

• Substitute each solution back into the original equation.
• Check that both sides of the equation are equal once the solution is substituted.

For example, in our act of verification:- Substituting \( x = 5 \) back into the equation \( 1 - \frac{3}{x} - \frac{10}{x^2} = 0 \) gives us \( 1 - \frac{3}{5} - \frac{2}{5} = 0 \), confirming \( x = 5 \) is indeed a solution.
- For \( x = -2 \), substituting back results in \( 1 + \frac{3}{2} - \frac{5}{2} = 0 \), so \( x = -2 \) is also a valid solution.This step ensures the accuracy of your answers and builds confidence in your problem-solving abilities.