Problem 48 - Algebra and Trigonometry

Question

Solve each equation in Exercises 41–60 by making an appropriate substitution. $$x^{-2}-x^{-1}-6=0$$

Step-by-Step Solution

Answer

The solutions of the given equation are \(x = 1/3\) and \(x = -1/2\).

Step 1: Make a substitution

Let's substitute \(u = x^{-1}\). Hence the equation will become \(u^2 - u - 6 = 0\). Now you have a simple quadratic equation.

Step 2: Solve the quadratic equation

To solve the quadratic equation \(u^2 - u - 6 = 0\), you first need to factorize it, which will yield \((u - 3)(u + 2) = 0\). Setting each factor equal to zero gives the solutions \(u = 3\) and \(u = -2\).

Step 3: Substitute back the original variable

Now that we have the solutions for \(u\), we need to substitute back \(u = x^{-1}\) to get the solutions for \(x\). Doing this, we yield \(x^{-1} = 3\) and \(x^{-1} = -2\). From these, we get \(x = 1/3\) and \(x = -1/2\) respectively.

Key Concepts

Substitution Method

Substitution is a technique used to solve equations more easily by replacing complex parts of an equation with simpler variables. This can make the problem more manageable by transforming a challenging equation into a form that is easier to handle.
For example, if you come across an equation with terms that involve higher powers or reciprocals, like in this exercise, you can create a substitution to simplify it. Replacing the complex term with a single variable can turn a complicated polynomial into a quadratic, which is much easier to solve.

• The first step is to identify the part of the equation that can be substituted for a new variable.
• Choose a substitution that leads to an equation with a solvable structure.
• After solving for the new variable, remember to substitute back to find the original variable’s value.

This technique is especially helpful when dealing with equations that might otherwise be challenging or tedious to solve.

Solving Equations

When solving equations, the main goal is to find the values that make the equation true. In general, this involves isolating the variable (or variables) of interest. With the substitution method, solving the equation becomes much more straightforward.
Consider the transformed equation from the substitution step: it becomes a simple quadratic equation, which is generally easier to solve than the original form. Techniques for solving include:

• Isolating the variable, if possible.
• Using inverse operations, such as squares and square roots, to simplify.
• For quadratics, using factoring, completing the square, or the quadratic formula.

Once you find the solutions to the simpler equation, you must then "solve backwards" to find the solutions in terms of the original variables. This ensures you find the correct answers that satisfy the initial equation's conditions.

Factoring Quadratics

Factoring is a widely used method for solving quadratic equations. It involves expressing the quadratic as a product of linear factors. For example, if you have an equation like \(u^2 - u - 6 = 0\), you look for two numbers that multiply to -6 (the constant term) and add up to -1 (the coefficient of \(u\)).

• In this case, these numbers are 2 and -3 because \((-3) \cdot 2 = -6\) and \(-3 + 2 = -1\).
• Then you write the quadratic as \((u - 3)(u + 2) = 0\).
• Solving these individual linear equations gives you the solutions \(u = 3\) and \(u = -2\).

Factoring is often preferred because it provides a quick way to get the solution without needing to use more complex methods like the quadratic formula. Understanding how to factor quickly and correctly is an essential skill in algebra that helps solve quadratic equations efficiently.