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Chapter 1: Problem 117

More on Solving Equations Find all real solutions of the equation. $$\frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0$$

Short Answer

Expert verified
The real solution is \(-\frac{1}{2}\).

Step by step solution

01

Identify a Common Factor

We notice that the equation \(\frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0\) has terms that all include powers of \(x\) in the denominator. A common strategy is to multiply through by the lowest power of \(x\) that clears the denominators, which in this case is \(x^3\).
02

Multiply Throughout by \(x^3\)

Multiply each term of the equation \(\frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0\) by \(x^3\) to eliminate the fractions:\[ x^3 \times \frac{1}{x^3} + x^3 \times \frac{4}{x^2} + x^3 \times \frac{4}{x} = 0 \] Simplifying gives:\[ 1 + 4x + 4x^2 = 0 \]
03

Rearrange as a Quadratic Equation

We now have a quadratic equation in terms of \(x\). Rearrange it as follows:\[ 4x^2 + 4x + 1 = 0 \]
04

Apply the Quadratic Formula

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a = 4\), \(b = 4\), and \(c = 1\), find the roots of the equation:Calculate the discriminant: \[ b^2 - 4ac = 4^2 - 4 \times 4 \times 1 = 16 - 16 = 0 \]Since the discriminant is zero, there is one real solution, given by:\[ x = \frac{-4 \pm \sqrt{0}}{2 \times 4} = \frac{-4}{8} = -\frac{1}{2} \]
05

Conclusion of Solutions

The only real solution from the quadratic equation is \(-\frac{1}{2}\). Hence, the set of real solutions for the original equation is \( \{-\frac{1}{2}\} \).

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Key Concepts

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

Quadratic Equations
Quadratic equations are a special form of polynomial equations of degree two. They have the structure \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). In the context of our original problem, multiplying through by \(x^3\) transforms it into \[ 4x^2 + 4x + 1 = 0 \]. This transformation is crucial because quadratic equations are well-studied, and we have powerful techniques to find their solutions.
  • Quadratic equations can often be factored into simpler expressions, but this is not always possible, especially when the equation has no real roots.
  • What makes quadratic equations very handy to deal with is that solutions can be found using the quadratic formula.
Understanding how we arrive at a quadratic form from something more complex, like the original equation, is key to leveraging these techniques.
Quadratic Formula
The quadratic formula is a universal tool for solving quadratic equations. It allows you to find the solutions (or roots) of any quadratic equation of the form \[ ax^2 + bx + c = 0 \] without necessarily factoring the equation. The formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula utilizes the coefficients \(a\), \(b\), and \(c\) from the equation.The key part of the quadratic formula is the discriminant, \(b^2 - 4ac\). It helps determine the nature of the roots:
  • If the discriminant is positive, there are two distinct real roots.
  • If the discriminant equals zero, there is exactly one real root.
  • If the discriminant is negative, the roots are complex or imaginary.
In our original equation, the discriminant was zero, hence there was one real solution.
Real Solutions
Real solutions of an equation are the solutions that do not involve any imaginary numbers. In our context, real solutions are particularly important because they are values of \(x\) that satisfy the equation in a real number sense.When working with quadratic equations, it's the discriminant (\(b^2 - 4ac\)) that tells us about the real solutions:
  • A zero discriminant means one real solution.
  • A positive discriminant indicates two different real solutions.
  • A negative discriminant means there are no real solutions, only complex ones.
In the example from our original problem, the discriminant was zero, giving one real solution, \(-\frac{1}{2}\). This means that the only value of \(x\) that makes the equation true is \(-\frac{1}{2}\), highlighting the use of the discriminant in determining real solutions.

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