Chapter 8: Problem 7
For each equation, determine the substitution that should be made to write the equation in quadratic form. a. \(x^{4}-12 x^{2}+27=0\) Let \(y=\) ___ b. \(x-13 \sqrt{x}+40=0\) Let \(y=\) ___ c. \(x^{2 / 3}+2 x^{1 / 3}-3=0\) Let \(y=\) ___ d. \(x^{-2}-x^{-1}-30=0\) Let \(y=\) ___ e. \((x+1)^{2}-(x+1)-6=0 \quad\) Let \(y=\) ___
Short Answer
Step by step solution
Identify the quadratic structure
Step 2a: Solve (a) for substitution
Step 2b: Solve (b) for substitution
Step 2c: Solve (c) for substitution
Step 2d: Solve (d) for substitution
Step 2e: Solve (e) for substitution
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Substitution Method
For example, consider the equation \(x^{4}-12x^{2}+27=0\). By identifying that the term \(x^{4}\) can be expressed as \((x^2)^2\), we can set \(y = x^{2}\). This substitution transforms the polynomial into \(y^{2} - 12y + 27 = 0\), a standard quadratic equation.
Key benefits of using the substitution method include:
- Reduction of higher-degree polynomials to quadratic equations, which are easier to solve.
- Simplifying the visual complexity of equations, making patterns more apparent.
- Facilitating the use of quadratic formula techniques to find solutions.
Polynomial Equations
A polynomial equation is generally of the form \(a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\), where \(a_n\) are coefficients and \(n\) is a positive integer. Solutions to polynomial equations refer to the variable values that satisfy the equation, setting it to zero. For instance, solving \((x+1)^2 - (x+1) - 6 = 0\) requires finding values of \(x\) that satisfy this expression.
In theoretical and applied math, polynomial equations are pivotal because:
- They form the basis for various algebraic structures.
- They are used to model real-world scenarios in physics, engineering, and economics.
- Understanding their roots helps in graphing polynomial functions and finding intercepts.
Negative Exponents
Taking the example equation \(x^{-2} - x^{-1} - 30 = 0\), negative exponents suggest a reciprocal relationship. Here, setting \(y = x^{-1}\) not only simplifies the terms but changes the equation into a recognizable quadratic format: \(y^2 - y - 30 = 0\).
Important points about negative exponents:
- They are often used to express fractions more succinctly.
- They can be applied to manipulate formulas in calculus and other advanced topics.
- Understanding them aids in solving a wider variety of algebraic equations effectively.
Rational Exponents
Consider the equation \(x^{2/3} + 2x^{1/3} - 3 = 0\). By observing the fractional powers, one can set a substitution like \(y = x^{1/3}\), transforming the equation into \(y^2 + 2y - 3 = 0\).
Benefits of understanding rational exponents include:
- Simplifying complex roots in equations to easily solvable forms.
- Applying algebraic techniques regardless of integer-based limitations.
- Enhancing clarity when dealing with polynomials that involve roots.