Problem 27 - A Graphical Approach to College Algebra

Question

Solve each equation by hand. Do not use a calculator. $$5 x^{-2}+13 x^{-1}=28$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{5}{7} \) and \( x = -\frac{1}{4} \).

Step 1: Substitute Variables

Let \( y = x^{-1} \). This means \( x^{-2} = y^2 \). Substituting these into the equation, \( 5x^{-2} + 13x^{-1} = 28 \) becomes \( 5y^2 + 13y = 28 \).

Step 2: Rearrange the Equation

Move all terms to one side of the equation to form a quadratic equation: \( 5y^2 + 13y - 28 = 0 \).

Step 3: Identify the Quadratic Formula

A quadratic equation in the form \( ay^2 + by + c = 0 \) can be solved using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \). Here, \( a = 5 \), \( b = 13 \), and \( c = -28 \).

Step 4: Calculate the Discriminant

Calculate the discriminant: \( b^2 - 4ac = 13^2 - 4(5)(-28) = 169 + 560 = 729 \).

Step 5: Solve for y

Substitute the values into the quadratic formula: \( y = \frac{-13 \pm \sqrt{729}}{10} \). Since \( \sqrt{729} = 27 \), solve for \( y \):- \( y = \frac{-13 + 27}{10} = \frac{14}{10} = 1.4 \)- \( y = \frac{-13 - 27}{10} = \frac{-40}{10} = -4 \).

Step 6: Substitute Back for x

Since \( y = x^{-1} \), we have \( x = \frac{1}{y} \), so- For \( y = 1.4 \), \( x = \frac{1}{1.4} = \frac{5}{7} \).- For \( y = -4 \), \( x = \frac{1}{-4} = -\frac{1}{4} \).

Key Concepts

Substitution Method

The substitution method simplifies complex equations by replacing them with terms that are easier to manage. In our exercise, we start by defining a new variable to make the equation less intimidating: let \( y = x^{-1} \). This transforms the original equation from a difficult form \( 5x^{-2} + 13x^{-1} = 28 \) into a simpler quadratic equation \( 5y^2 + 13y = 28 \). By making such substitutions, we reduce the algebraic complexity, allowing us to solve a different but equivalent equation with more ease.

• Step One: Identify the substitution—here, consider the original roles of the powers of \( x \).
• Step Two: Translate the entire equation using the substitution.
• Step Three: Solve the new equation to find the substituted variable.

Please note: After finding the values for the substituted variable, you must translate back to the original variable to complete the solution.

Quadratic Formula

The quadratic formula is a vital tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). By using this formula, we avoid the need for complex factorization. For the equation \( 5y^2 + 13y - 28 = 0 \), we identify \( a = 5 \), \( b = 13 \), and \( c = -28 \). The quadratic formula is:\[ y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
This equation gives us the solution for \( y \) in one step. By plugging in the values of \( a \), \( b \), and \( c \), we find our solutions for \( y \). This formula streamlines the process greatly, providing a reliable method for solving any quadratic equation as long as it fits this standard form.

• Step One: Plug the identified \( a \), \( b \), and \( c \) into the quadratic formula.
• Step Two: Calculate the part under the square root, known as the discriminant.
• Step Three: Compute the values of \( y \) using both the plus and minus aspects of the formula.

Discriminant Calculation

The discriminant is a crucial component of the quadratic formula, found under the square root symbol: \( b^2 - 4ac \). This calculation not only helps in solving the quadratic but also provides insight into the nature of the roots.
For our equation, the discriminant is calculated as:\[ 13^2 - 4 \times 5 \times (-28) = 169 + 560 = 729 \]
A positive discriminant indicates two real and distinct solutions, which was observed in our example since 729 is a positive number. Conversely, a discriminant of zero suggests one real solution, while a negative discriminant points to complex roots, which we don't encounter here.

• Positive Discriminant: Two different real solutions.
• Zero Discriminant: One repeated real solution.
• Negative Discriminant: No real solutions, but complex numbers instead.

Understanding the discriminant helps in predicting the type of solutions even before completely solving the equation.