Simplify each expression. Write answers without negative exponents. Assume all vari-ables represent positive real numbers. See Examples 8 and 9. (p1/5p7/10p1/2)/(p3)-1/5
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
0. Review of Algebra
Rational Exponents
Problem 144
Textbook Question
Solve for x: x^(5/6) + x^(2/3) - 2x^(1/2) = 0
Verified step by step guidance1
Rewrite the exponents in terms of a common base. Notice that all the exponents are fractions with a denominator of 6. Rewrite the terms as follows: \(x^{5/6} = (x^{1/6})^5\), \(x^{2/3} = (x^{1/6})^4\), and \(x^{1/2} = (x^{1/6})^3\).
Let \(y = x^{1/6}\). This substitution simplifies the equation to \(y^5 + y^4 - 2y^3 = 0\).
Factor the simplified equation. Start by factoring out the greatest common factor, \(y^3\), which gives \(y^3(y^2 + y - 2) = 0\).
Solve for \(y\). The equation \(y^3 = 0\) gives \(y = 0\). For the quadratic \(y^2 + y - 2 = 0\), factor it as \((y + 2)(y - 1) = 0\), which gives \(y = -2\) and \(y = 1\).
Substitute back \(y = x^{1/6}\) into each solution. For \(y = 0\), \(x^{1/6} = 0\), so \(x = 0\). For \(y = 1\), \(x^{1/6} = 1\), so \(x = 1\). Discard \(y = -2\) because \(x^{1/6}\) cannot be negative for real numbers. The solutions are \(x = 0\) and \(x = 1\).
Verified video answer for a similar problem:This video solution was recommended by our tutors as helpful for the problem above
Video duration:
13mPlay a video:
Was this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Exponents and Fractional Powers
Understanding exponents, particularly fractional powers, is crucial in this equation. The expression x^(5/6) indicates that x is raised to the power of 5/6, which can be interpreted as the sixth root of x raised to the fifth power. This concept is essential for manipulating and simplifying the equation effectively.
Recommended video:
Powers of i
Factoring Polynomials
Factoring is a key technique in solving polynomial equations. In this case, the equation can be rearranged and factored to find the values of x that satisfy the equation. Recognizing common factors or using substitution can simplify the process of finding solutions.
Recommended video:
Guided course
Introduction to Factoring Polynomials
Roots of Equations
Finding the roots of an equation involves determining the values of x that make the equation equal to zero. This concept is fundamental in algebra, as it allows us to identify solutions to polynomial equations. Techniques such as the Rational Root Theorem or numerical methods may be employed to find these roots.
Recommended video:
Solving Quadratic Equations by the Square Root Property
Related Videos
Related Practice
Textbook Question
443
views
