Problem 17 - A Graphical Approach to College Algebra

Question

Find all complex solutions for each equation by hand. Do not use a calculator. $$1-\frac{13}{x}+\frac{36}{x^{2}}=0$$

Step-by-Step Solution

Answer

The solutions are \( x = 9 \) and \( x = 4 \).

Step 1: Rewrite the Equation

To solve the equation, let's rewrite it in a more standard form. Multiply every term in the equation by \( x^2 \) to eliminate the fractions:\[1x^2 - 13x + 36 = 0\]Now, we have a quadratic equation in the standard form \( ax^2 + bx + c = 0 \) where \( a = 1 \), \( b = -13 \), and \( c = 36 \).

Step 2: Calculate the Discriminant

The next step is to calculate the discriminant \( \Delta \) of the quadratic equation, which is given by \( \Delta = b^2 - 4ac \). Using our coefficients, we calculate:\[\Delta = (-13)^2 - 4 \times 1 \times 36 = 169 - 144 = 25\]Since the discriminant is positive, we know there are two distinct real solutions.

Step 3: Apply the Quadratic Formula

Now, use the quadratic formula to find the solutions to the quadratic equation:\[x = \frac{-b \pm \sqrt{\Delta}}{2a}\]Substitute the known values into the formula:\[x = \frac{-(-13) \pm \sqrt{25}}{2 \times 1} = \frac{13 \pm 5}{2}\]

Step 4: Solve for the Roots

Calculate the two possible values for \( x \):1. \( x = \frac{13 + 5}{2} = \frac{18}{2} = 9 \)2. \( x = \frac{13 - 5}{2} = \frac{8}{2} = 4 \) Thus, the solutions are \( x = 9 \) and \( x = 4 \).

Step 5: Verify the Solutions

Substitute the solutions back into the original equation to verify.For \( x = 9 \):\[1 - \frac{13}{9} + \frac{36}{9^2} = 0 \1 - \frac{13}{9} + \frac{36}{81} = 0 \1 - \frac{13}{9} + \frac{4}{9} = 0 \\frac{9}{9} - \frac{13}{9} + \frac{4}{9} = 0 \-\frac{9}{9} = 0 \checkmark\]For \( x = 4 \):\[1 - \frac{13}{4} + \frac{36}{4^2} = 0 \1 - \frac{13}{4} + \frac{36}{16} = 0 \1 - \frac{13}{4} + \frac{9}{4} = 0 \\frac{4}{4} - \frac{13}{4} + \frac{9}{4} = 0 \-\frac{4}{4} = 0 \checkmark\]Both solutions satisfy the original equation.

Key Concepts

Quadratic Equation

A quadratic equation is a second-order polynomial equation in a single variable, usually in the form of \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants with \( a eq 0 \). The term "quadratic" comes from "quad" meaning "square" because the variable \( x \) is squared in the equation.

Quadratic equations are used to find the values of \( x \) for which the equation holds true. These values are known as the roots or solutions of the equation. Quadratic equations can have:

• Two distinct real roots
• One real double root
• Two complex conjugate roots

The types of solutions depend on the discriminant of the quadratic equation, which we'll explore shortly.

Discriminant

In the context of quadratic equations, the discriminant is a key component that determines the nature of the roots of the equation. The discriminant, usually denoted by \( \Delta \), is calculated from the equation's coefficients \( a \), \( b \), and \( c \) using the formula:

\[ \Delta = b^2 - 4ac \]

The discriminant provides insights into the number and type of solutions, such as whether they are real or complex. Here's how you can interpret the discriminant:

• \( \Delta > 0 \): Two distinct real roots
• \( \Delta = 0 \): One real repeated root
• \( \Delta < 0 \): Two complex conjugate roots

Calculating the discriminant is an essential step when solving quadratic equations, as it tells you what to expect from the roots before computing them using the quadratic formula.

Quadratic Formula

The quadratic formula is a universally accepted formula that provides the solutions to a quadratic equation. It is derived from the process of completing the square and works for any quadratic equation in the standard form. The quadratic formula is expressed as:

\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \]

where \( b \) and \( a \) are the coefficients of the equation and \( \Delta \) is the discriminant. To find the solutions:

• Calculate the discriminant \( \Delta = b^2 - 4ac \)
• Substitute \( a \), \( b \), and \( \sqrt{\Delta} \) into the quadratic formula
• Compute the values of \( x \) using both the plus \( + \) and minus \( - \) options

By substituting the values into this formula, you can obtain up to two roots, which could be real or complex depending on the discriminant. The quadratic formula is a powerful tool because it provides a straightforward method to solve any quadratic equation without needing to factor or complete the square.