Problem 42
Question
For the following exercises, solve for the unknown variable. $$ x^{-2}-x^{-1}-12=0 $$
Step-by-Step Solution
Verified Answer
The solutions are \(x = \frac{1}{4}\) and \(x = -\frac{1}{3}\).
1Step 1: Identify the Problem Type
The given expression is a quadratic equation in terms of powers of \(x\), specifically involving \(x^{-2}\) and \(x^{-1}\). Our job is to solve for the unknown variable \(x\).
2Step 2: Substitute with a New Variable
Let \(y = x^{-1}\). This means \(x^{-2} = y^2\). By substituting, the equation becomes \(y^2 - y - 12 = 0\). This is a standard quadratic equation.
3Step 3: Solve the Quadratic Equation
Factor the equation \(y^2 - y - 12 = 0\). It factors as \((y - 4)(y + 3) = 0\). Therefore, the solutions for \(y\) are \(y = 4\) and \(y = -3\).
4Step 4: Substitute Back to Find x
Since \(y = x^{-1}\), this implies \(x = y^{-1}\). When \(y = 4\), \(x = \frac{1}{4}\). When \(y = -3\), \(x = -\frac{1}{3}\).
5Step 5: Verify the Solutions
Substitute \(x = \frac{1}{4}\) back into the original equation to verify: \((\frac{1}{4})^{-2} - (\frac{1}{4})^{-1} - 12 = 16 - 4 - 12 = 0\). Next, substitute \(x = -\frac{1}{3}\) to verify: \((-\frac{1}{3})^{-2} - (-\frac{1}{3})^{-1} - 12 = 9 + 3 - 12 = 0\). Both values satisfy the original equation.
Key Concepts
Variables SubstitutionFactoring QuadraticsInverse Functions
Variables Substitution
In mathematics, sometimes expressions can be simplified, or more easily understood, by substituting a part of them with a new variable. This process is known as "variable substitution". It involves choosing a complex part of an equation and replacing it with a simpler symbol to make calculations easier.
The original problem presented is an equation involving powers of a variable, specifically terms like \( x^{-2} \) and \( x^{-1} \). To solve the equation, we can substitute \( x^{-1} \) with the simpler variable \( y \). Thus, \( x^{-2} \) becomes \( y^2 \). Now, we transform the original equation into \( y^2 - y - 12 = 0 \), simplifying our work considerably.
This substitution converts the given equation into a quadratic form, which can be solved using familiar methods. Essentially, by replacing complex parts with simpler variables, calculations become less daunting, and understanding the problem can be straightforward.The key reasons for variable substitution include:
The original problem presented is an equation involving powers of a variable, specifically terms like \( x^{-2} \) and \( x^{-1} \). To solve the equation, we can substitute \( x^{-1} \) with the simpler variable \( y \). Thus, \( x^{-2} \) becomes \( y^2 \). Now, we transform the original equation into \( y^2 - y - 12 = 0 \), simplifying our work considerably.
This substitution converts the given equation into a quadratic form, which can be solved using familiar methods. Essentially, by replacing complex parts with simpler variables, calculations become less daunting, and understanding the problem can be straightforward.The key reasons for variable substitution include:
- Simplifying complex expressions.
- Making equations more familiar and easier to solve.
- Reducing calculation errors.
Factoring Quadratics
Once we have rewritten a complex equation as a standard quadratic equation, the next step is often to solve it by factoring. A quadratic equation of the form \( ax^2 + bx + c = 0 \) can sometimes be expressed as the product of two simpler expressions. Factoring involves finding these expressions that multiply to give the original quadratic.
In the given exercise, after substitution, we derived \( y^2 - y - 12 = 0 \). The goal is to express this in the form \((y - p)(y + q) = 0\), where \( p \) and \( q \) are numbers whose product equals \(-12\) and whose sum equals \(-1\) (the coefficient of \( y \)).
The equation factors neatly into \((y - 4)(y + 3) = 0\). This implies that either \( y - 4 = 0 \) or \( y + 3 = 0 \). From which we deduce the solutions for \( y \) are 4 and -3.
Factoring quadratics allows us to break down complex equations into simple, solvable parts. It uses relationships between numbers and their properties, making it a powerful tool for solving quadratic problems effectively.
In the given exercise, after substitution, we derived \( y^2 - y - 12 = 0 \). The goal is to express this in the form \((y - p)(y + q) = 0\), where \( p \) and \( q \) are numbers whose product equals \(-12\) and whose sum equals \(-1\) (the coefficient of \( y \)).
The equation factors neatly into \((y - 4)(y + 3) = 0\). This implies that either \( y - 4 = 0 \) or \( y + 3 = 0 \). From which we deduce the solutions for \( y \) are 4 and -3.
Factoring quadratics allows us to break down complex equations into simple, solvable parts. It uses relationships between numbers and their properties, making it a powerful tool for solving quadratic problems effectively.
Inverse Functions
Inverse functions are essential when working with equations involving inverses or reciprocal values. An inverse of a function essentially "undoes" the action of the original function. For example, if \( y = x^{-1} \), then \( x = y^{-1} \).
In this context, after solving the quadratic in the substituted variable \( y \), the solutions \( y = 4 \) and \( y = -3 \) translate back to \( x \) using inverse functions. Specifically, since \( y = x^{-1} \), the solution for \( x \) becomes \( x = 4^{-1} = \frac{1}{4} \) and \( x = (-3)^{-1} = -\frac{1}{3} \).
This step is crucial in returning to the original variable of the problem and checking that our found solutions satisfy the initial equation. Understanding inverse relationships helps in effectively translating between different forms of equations and in solving real-world reciprocal problems.
In this context, after solving the quadratic in the substituted variable \( y \), the solutions \( y = 4 \) and \( y = -3 \) translate back to \( x \) using inverse functions. Specifically, since \( y = x^{-1} \), the solution for \( x \) becomes \( x = 4^{-1} = \frac{1}{4} \) and \( x = (-3)^{-1} = -\frac{1}{3} \).
This step is crucial in returning to the original variable of the problem and checking that our found solutions satisfy the initial equation. Understanding inverse relationships helps in effectively translating between different forms of equations and in solving real-world reciprocal problems.
Other exercises in this chapter
Problem 41
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