Find all values of x satisfying the given conditions. y1 = 2x2 + 5x - 4, y2 = - x2 + 15x - 10, and y1 - y2 = 0
Table of contents
- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
1. Equations & Inequalities
Choosing a Method to Solve Quadratics
Problem 52
Textbook Question
Solve each equation in Exercises 41–60 by making an appropriate substitution.
Verified step by step guidance1
Identify the substitution to simplify the equation. Notice that the exponents are fractional and related: \(x^{2/5}\) and \(x^{1/5}\). Let \(u = x^{1/5}\), so that \(u^2 = x^{2/5}\).
Rewrite the original equation in terms of \(u\): replace \(x^{2/5}\) with \(u^2\) and \(x^{1/5}\) with \(u\). The equation becomes \(u^2 + u - 6 = 0\).
Solve the quadratic equation \(u^2 + u - 6 = 0\) using factoring, completing the square, or the quadratic formula. This will give you the possible values of \(u\).
After finding the values of \(u\), recall that \(u = x^{1/5}\). To find \(x\), raise both sides of the equation \(u = x^{1/5}\) to the 5th power, resulting in \(x = u^5\).
Calculate \(x\) for each value of \(u\) found in step 3 by computing \(x = u^5\). These are the solutions to the original equation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Exponents and Rational Powers
Rational exponents represent roots and powers, such as x^(m/n) meaning the n-th root of x raised to the m-th power. Understanding how to manipulate and simplify expressions with fractional exponents is essential for rewriting and solving equations involving these terms.
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Substitution Method
Substitution involves replacing a complex expression with a simpler variable to transform the equation into a more familiar form, often a polynomial. This technique simplifies solving equations that contain complicated terms, such as fractional powers, by reducing them to quadratic or linear equations.
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Solving Quadratic Equations
Once substitution reduces the equation to a quadratic form, solving it involves finding values of the variable that satisfy the equation using factoring, completing the square, or the quadratic formula. Understanding how to solve quadratics is crucial to find the solutions to the original equation after back-substitution.
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