Problem 76 - Algebra A Combined Function

Question

Solve. $$ y^{-2}-8 y^{-1}+7=0 $$

Step-by-Step Solution

Answer

The solutions for \( y \) are \( y = \frac{1}{7} \) and \( y = 1 \).

Step 1: Substitute the Variable

Let's set a substitution to make the equation easier to solve. Let \( z = y^{-1} \), meaning \( y^{-2} = z^2 \). The original equation \( y^{-2} - 8y^{-1} + 7 = 0 \) becomes \( z^2 - 8z + 7 = 0 \).

Step 2: Solve the Quadratic Equation

The equation \( z^2 - 8z + 7 = 0 \) is a quadratic equation in \( z \). We can solve it using the quadratic formula: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1 \), \( b = -8 \), \( c = 7 \). This gives: \[ z = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 7}}{2 \cdot 1} \] \[ z = \frac{8 \pm \sqrt{64 - 28}}{2} \] \[ z = \frac{8 \pm \sqrt{36}}{2} \] \[ z = \frac{8 \pm 6}{2} \]

Step 3: Calculate the Solutions for z

Continuing from the quadratic formula outputs: 1. \( z = \frac{8 + 6}{2} = 7 \)2. \( z = \frac{8 - 6}{2} = 1 \) Thus, the solutions for \( z \) are \( z = 7 \) and \( z = 1 \).

Step 4: Convert Solutions Back to y

Since we set \( z = y^{-1} \) in Step 1, we can find \( y \) by taking the reciprocal of \( z \).For \( z = 7 \), we have \( y = \frac{1}{7} \).For \( z = 1 \), we have \( y = 1 \).

Key Concepts

Substitution Method

The substitution method is a powerful technique to simplify complex equations by replacing variables with simpler expressions. This often transforms intricate equations into more manageable forms. In solving equations like the given one, \( y^{-2} - 8y^{-1} + 7 = 0 \), the substitution method involves changing the variable to something much simpler.

By letting \( z = y^{-1} \), we directly set \( y^{-2} = z^2 \). This substitution transforms the original problem into a more familiar quadratic equation \( z^2 - 8z + 7 = 0 \). The substitution method simplifies different types of equations, making it easier to apply other solving techniques like factoring or using the quadratic formula.

• Simplification: Convert complex powers to simple quadratic expressions.
• Ease of solving: Facilitates use of known techniques such as the quadratic formula.
• Flexibility: Applicable to various types of equations when direct solution is challenging.

Once the quadratic form is solved, reverse the substitution to find the original variable. This approach not only solves the problem but also illustrates how transformations can simplify otherwise daunting tasks.

Quadratic Formula

The quadratic formula is a universal tool to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). In the context of our problem, with the equation \( z^2 - 8z + 7 = 0 \), the quadratic formula is applied to find the solutions for \( z \).

The formula is: \[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where in our equation, \( a = 1 \), \( b = -8 \), and \( c = 7 \). Plugging these values into the formula allows us to solve for \( z \) directly:

• Calculate the discriminant: \( b^2 - 4ac = (-8)^2 - 4 \times 1 \times 7 = 64 - 28 = 36 \).
• Determine the square root of the discriminant \( \sqrt{36} = 6 \).
• Substitute into the formula: \( z = \frac{8 \pm 6}{2} \) gives two possible solutions.

These calculations yield \( z = 7 \) and \( z = 1 \). Understanding how to use the quadratic formula is essential for solving quadratic equations efficiently and accurately. It provides a structured approach that works every time, regardless of whether the equation can easily be factored or not.

Reciprocal Functions

Reciprocal functions involve taking the reciprocal of a variable or number, which effectively means "flipping" the value. If you have a number \( x \), its reciprocal is \( \frac{1}{x} \). In this exercise, after finding the value of \( z \) using the quadratic formula, the final step involves finding the original variable \( y \) by taking the reciprocal of \( z \).

In other words, since we earlier set \( z = y^{-1} \), solving for \( y \) involves:

• For \( z = 7 \), the corresponding \( y \) is \( \frac{1}{7} \).
• For \( z = 1 \), the corresponding \( y \) is \( 1 \).

Reciprocal functions often appear in equations where inverses or transformations are required. They are crucial in mathematics because they offer insights into relationships between numbers and their inverses, showing how changes to one part of an equation reflect through the whole. Understanding reciprocals is key to reversing substitutions and arriving at solutions to transformed equations.