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Chapter 5: Problem 31

Solve each equation by hand. Do not use a calculator. $$x^{3 / 4}-x^{1 / 2}-x^{1 / 4}+1=0$$

Short Answer

Expert verified
The solution is \( x = 1 \).

Step by step solution

01

Identify Like Terms

We notice that the equation \( x^{3/4} - x^{1/2} - x^{1/4} + 1 = 0 \) can be simplified by recognizing like terms which have the same base with different exponents.
02

Substitute Variables

To simplify the equation, let's use the substitution: \( y = x^{1/4} \). Then we rewrite the equation in terms of \( y \): \( y^3 - y^2 - y + 1 = 0 \).
03

Solve the Simplified Equation

The equation \( y^3 - y^2 - y + 1 = 0 \) is a cubic equation. We can try to find rational roots using the Rational Root Theorem.
04

Rational Root Theorem

By the Rational Root Theorem, possible rational roots are \( \pm 1 \). Testing these, \( y = 1: (1)^3 - (1)^2 - 1 + 1 = 0 \), so \( y = 1 \) is a root.
05

Factor the Cubic Equation

Since \( y = 1 \) is a root, \( y-1 \) is a factor. We perform polynomial division of \( y^3 - y^2 - y + 1 \) by \( y-1 \), obtaining \( (y - 1)(y^2 + 0y - 1) = 0 \).
06

Solve the Quadratic Equation

Now solve \( y^2 - 1 = 0 \), which factors to \( (y-1)(y+1) = 0 \). Thus, \( y = 1 \) or \( y = -1 \).
07

Back-Substitute to \( x \)

Since \( y = x^{1/4} \), substitute back: \( x^{1/4} = 1 \) or \( x^{1/4} = -1 \). Solve: \( x = 1^4 = 1 \) or \( x = (-1)^4 = 1 \).
08

Check the Solution

Substitute \( x = 1 \) into the original equation to verify: \( 1^{3/4} - 1^{1/2} - 1^{1/4} + 1 = 0 \), which holds true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Root Theorem
Understanding the Rational Root Theorem is a powerful tool when dealing with polynomial equations. This theorem helps us identify potential rational solutions for polynomial equations with integer coefficients. Here's the essence:
  • Consider a polynomial: \( a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0 \) with integer coefficients.
  • The Rational Root Theorem states that any rational root, written as \( \frac{p}{q} \), must satisfy that \( p \) is a factor of the constant term \( a_0 \), and \( q \) is a factor of the leading coefficient \( a_n \).
  • This theorem narrows down the possibilities for rational roots, making the search process more manageable.
In our problem, once the cubic equation \( y^3 - y^2 - y + 1 = 0 \) is derived, the Rational Root Theorem suggests that \( \pm 1 \) are possible roots. Testing these values guides us toward discovering that \( y = 1 \) indeed satisfies the equation. By using this theorem, identifying plausible solutions becomes quicker and more efficient.
Substitution Method
The substitution method is a simplifying technique that makes solving complicated expressions more manageable. In our exercise, we face an equation with fractional exponents, which initially seems daunting. Here鈥檚 how substitution helps:
  • We have an equation: \( x^{3/4} - x^{1/2} - x^{1/4} + 1 = 0 \).
  • Introduce a substitution: Let \( y = x^{1/4} \).
  • This converts fractional exponents into integers, simplifying our work. The equation transforms into: \( y^3 - y^2 - y + 1 = 0 \).
Using substitution simplifies the equation, maintaining the essence of the problem but in a more digestible form. This method allows us to apply familiar polynomial techniques, streamlining the solution process. Substitution transforms the original problem into a cubic equation, which is easier to tackle and solve.
Polynomial Division
Polynomial division is similar to basic arithmetic division but applied to polynomials. When solving cubic equations, especially after finding a root, polynomial division extracts remaining factors from the polynomial. Let's walk through the division process here:
  • Start with the given polynomial: \( y^3 - y^2 - y + 1 \).
  • Upon finding \( y = 1 \) as a root, know that \( y - 1 \) is a factor.
  • Perform polynomial division: Divide \( y^3 - y^2 - y + 1 \) by \( y - 1 \).
Through polynomial long division or synthetic division, the operation reveals another polynomial, typically a simpler one. In this case, we find \( (y - 1)(y^2 - 1) \), indicating how the original cubic equation factors into simpler components. This step is crucial for breaking down complex polynomials into manageable pieces.
Quadratic Equations
Quadratic equations are central in mathematics, representing the polynomial equation: \( ax^2 + bx + c = 0 \). In our steps, we encounter a quadratic as part of the factorization process. Here's how it fits in:
  • After factoring the cubic equation, we obtained: \( (y - 1)(y^2 - 1) = 0 \).
  • The quadratic: \( y^2 - 1 \) is equivalent to \( (y - 1)(y + 1) \).
  • We solve this quadratic equation: Potential solutions are found as \( y = 1 \) and \( y = -1 \).
Quadratics are straightforward to solve due to their structure, offering real or complex solutions depending on discriminant values. In our example, we factor and find real roots, guiding us to solve back for \( x \). Solving quadratics effectively bridges between finding initial roots and deeply analyzing further solutions.

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Most popular questions from this chapter

Solve the equation in part (a) graphically, expressing solutions to the nearest hundredth. Then use the graph to solve the associated inequalities in parts (b) and (c), expressing endpoints to the nearest hundredth. (a) \(\frac{\sqrt{2} x+5}{x^{3}-\sqrt{3}}=0\) (b) \(\frac{\sqrt{2} x+5}{x^{3}-\sqrt{3}}>0\) (c) \(\frac{\sqrt{2} x+5}{x^{3}-\sqrt{3}}<0\)

Solve each equation and inequality. (These types of equations and inequalities occur in calculus.) (a) \(\frac{(x-1)(2 x)-\left(x^{2}\right)(1)}{(x-1)^{2}}=0\) (b) \(\frac{(x-1)(2 x)-\left(x^{2}\right)(1)}{(x-1)^{2}}>0\)

CONCEPT CHECK In some cases, it is possible to solve a rational inequality simply by deciding what sign the numerator and the denominator must have and then using the rules for quotients of positive and negative numbers to determine the solution set. For example, consider the rational inequality $$ \frac{1}{x^{2}+1}>0 $$ The numerator of the rational expression, 1, is positive, and the denominator, \(x^{2}+1,\) must always be positive because it is the sum of a nonnegative number, \(x^{2},\) and a positive number, 1. Therefore, the rational expression is the quotient of two positive numbers, which is positive. Because the inequality requires that the rational expression be greater than \(0,\) and this will always be true, the solution set is \((-\infty, \infty)\) Use similar reasoning to solve each inequality. $$\frac{x^{4}+2}{-6} \geq 0$$

Solve each rational inequality by hand. $$2-\frac{5}{x}+\frac{2}{x^{2}} \geq 0$$

Use a calculator to graph rational function in the window indicated. Then (a) give the \(x\) - and y-intercepts, (b) explain why there are no vertical asymptotes, (c) give the equation of the oblique asymptote, and (d) give the domain and range. $$f(x)=\frac{x^{3}+4 x^{2}-x-4}{-2 x^{2}-2 x-4} ;[-6.6,6.6] \text { by }[-4.1,4.1]$$
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