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Magazine Chapter 8: Problem 42
Solve each equation. \(2 x^{-2}-5 x^{-1}-3=0\)
Short Answer
Expert verified
The solutions for the equation are \(x = \frac{1}{3}\) and \(x = -2\).
Step by step solution
01
Rewrite the Equation
Change all the terms involving the variable to a common form. By noticing that we have terms involving \(x^{-2}\) and \(x^{-1}\), introduce a new variable, \(y = x^{-1}\), so that \(x^{-2} = y^2\). Substitute these into the equation, getting:\[2y^2 - 5y - 3 = 0\]
02
Identify the Quadratic Equation
Recognize the equation from Step 1 as a quadratic equation in terms of \(y\):\[2y^2 - 5y - 3 = 0\]
03
Apply the Quadratic Formula
Use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -5\), and \(c = -3\) to solve for \(y\):\[y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2}\]Simplify the equation:\[y = \frac{5 \pm \sqrt{25 + 24}}{4}\]\[y = \frac{5 \pm \sqrt{49}}{4}\]\[y = \frac{5 \pm 7}{4}\]
04
Calculate the Solutions for y
Calculate the values of \(y\) from the possible solutions:1. \(y_1 = \frac{5 + 7}{4} = 3\)2. \(y_2 = \frac{5 - 7}{4} = -\frac{1}{2}\)
05
Convert Back to Solve for x
Substitute back using \(y = x^{-1}\) to find \(x\):1. When \(y = 3\), \(3 = x^{-1}\), so \(x = \frac{1}{3}\).2. When \(y = -\frac{1}{2}\), \(-\frac{1}{2} = x^{-1}\), so \(x = -2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a reliable tool for solving quadratic equations, which are equations in the form of \[ ax^2 + bx + c = 0. \] It's a vital method because it works for any quadratic equation, no matter the values of the coefficients. The formula is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \] Here鈥檚 a breakdown of how it works.
- **Identify** the coefficients: In the standard form of a quadratic equation, identify the values of \(a\), \(b\), and \(c\) from the equation. For example, if given \(2y^2 - 5y - 3 = 0\), \(a\) is 2, \(b\) is -5, and \(c\) is -3.
- **Calculate the Discriminant**: The discriminant \(b^2 - 4ac\) under the square root sign tells us about the nature of the roots:
- If it鈥檚 positive, the quadratic has two real and distinct solutions.
- If it鈥檚 zero, there is one real solution.
- If it鈥檚 negative, there are no real solutions but two complex solutions.
- **Solve for the roots**: Substitute \(b^2 - 4ac\) into the formula to find \(x\). Consider both the plus and minus cases to get both solutions.
Substitution Method
The substitution method is a technique used to simplify equations, especially when dealing with complex or high-order terms. When applied effectively, it converts an intricate problem into a standard algebraic form, making it easier to solve. In the provided exercise, we use substitution to solve the equation containing negative exponents.
To apply substitution:
To apply substitution:
- **Identify complex terms**: In equations like \(2x^{-2} - 5x^{-1} - 3 = 0\), look for expressions like \(x^{-2}\) and \(x^{-1}\) which are cumbersome to deal with directly.
- **Introduce a new variable**: Assign a new variable such as \(y = x^{-1}\) to express parts of the equation in a more manageable form. In our case, \(x^{-2}\) becomes \(y^2\).
- **Rewrite the equation**: Substitute these new terms back into the original equation, converting it into a simpler quadratic equation like \(2y^2 - 5y - 3 = 0\).
Inverse Operations
Inverse operations are a fundamental concept in algebra used to reverse operations to solve equations and isolate variables. When you have solved for a different variable using substitution, inverse operations allow you to wrangle the equation back to its original variables.
Here's how inverse operations work in solving our exercise:
Here's how inverse operations work in solving our exercise:
- **Understand inverse relationships**: If \(y = x^{-1}\), then \(x = y^{-1}\), meaning these operations undo each other.
- **Apply in context**: Once a solution is found in terms of the substitute variable, as in obtaining \(y = 3\) or \(y = -\frac{1}{2}\), find \(x\) by calculating the inverse, because \(x = \frac{1}{y}\). When \(y = 3\), \(x = \frac{1}{3}\) and when \(y = -\frac{1}{2}\), \(x = -2\).
