Problem 47 - College Algebra Essentials

Question

By making an appropriate substitution. $$ x^{-2}-x^{-1}-20=0 $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x = 1/5\) and \(x = -1/4\)

Step 1: Substitution

First, make a substitution for \(x^{-1}\), substituting with \(z\), so we have: \(z^{2}-z-20=0\)

Step 2: Applying Quadratic formula

Our equation now looks like a standard quadratic equation, which can be solved using the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \[a=1, b=-1, c=-20\]. Hence, \(z = \frac{-(-1) \pm \sqrt{(-1)^2 - 4*1*(-20)}}{2*1} = \frac{1 \pm \sqrt{1+80}}{2} = \frac{1 \pm \sqrt{81}}{2} = \frac{1 \pm 9}{2} \)

Step 3: Result for Substitution

On solving the above we get two solutions for \(z\): 5 and -4.

Step 4: Re-substitution

We originally set \(z = x^{-1}\), so \(x = 1 / z\). Hence, \(x = 1/5\) and \(x = -1/4\).

Key Concepts

Substitution Method

Understanding the substitution method is crucial when dealing with more complicated algebraic expressions, especially when we encounter quadratic equations in different formats. This method involves replacing a complex part of the equation with a simpler variable to make the equation easier to solve.

In the exercise given, the substitution method is used to transform a seemingly non-standard quadratic equation involving negative exponents into a recognizable quadratic form. Here's how it's done: The term with the highest negative exponent, in this case, x^{-1}, is assigned a new variable, let's say z. The original equation, which appears as x^{-2} - x^{-1} - 20 = 0, becomes z^2 - z - 20 = 0 when substituted with z = x^{-1}.

By applying this method, the equation is simplified and can now be approached with techniques applicable to standard quadratic equations.

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form ax^2 + bx + c = 0. The formula is z = (-b ± √(b^2 - 4ac)) / (2a). It provides the roots of the quadratic equation by substituting the coefficients a, b, and c into the formula.

In the exercise, after the substitution is made, the coefficients identified are a = 1, b = -1, and c = -20. Plugging these into the quadratic formula will give us the solutions for z, which are later re-substituted to find the actual values for x. It's an elegant process that guarantees a solution for any quadratic equation, provided that it has real roots.

Inverse Operations

Inverse operations are used to solve equations by reversing operations to isolate the variable. In the context of the provided exercise, once the substitute variable z has been solved using the quadratic formula, we must perform an inverse operation to revert back to the original variable x.

Since initially, we set z = x^{-1}, the inverse operation to 'undo' this is to take the reciprocal. Therefore, if z = 5 or z = -4, the inverse operations give us x = 1/5 and x = -1/4 respectively. Utilizing inverse operations is a vital step in ensuring that the solution is presented in terms of the original variables posed in the question.