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Chapter 1: Problem 88

Find all real solutions of the equation. $$\frac{1}{x-1}-\frac{2}{x^{2}}=0$$

Short Answer

Expert verified
There are no real solutions.

Step by step solution

01

Understand the equation

The given equation is \( \frac{1}{x-1} - \frac{2}{x^2} = 0 \). To solve it, our goal is to find the values of \( x \) that make the equation true (i.e., the expressions on either side of the equation are equal).
02

Set up a common denominator

To simplify the equation, find a common denominator. The common denominator for \( \frac{1}{x-1} \) and \( \frac{2}{x^2} \) is \( x^2(x-1) \).
03

Rewrite each fraction with the common denominator

Express each term with the common denominator:\[\frac{1}{x-1} = \frac{x^2}{x^2(x-1)}, \quad \frac{2}{x^2} = \frac{2(x-1)}{x^2(x-1)}\]
04

Combine the fractions

Combine the fractions over the common denominator:\[\frac{x^2}{x^2(x-1)} - \frac{2(x-1)}{x^2(x-1)} = \frac{x^2 - 2(x-1)}{x^2(x-1)}\]
05

Simplify the numerator

Simplify the numerator by expanding and combining like terms:\[x^2 - 2(x-1) = x^2 - 2x + 2\]
06

Solve the equation

Set the simplified numerator equal to zero to solve for \( x \):\[x^2 - 2x + 2 = 0\]This is a quadratic equation. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \( a=1, b=-2, c=2 \).
07

Calculate the discriminant

Compute the discriminant \( b^2 - 4ac = (-2)^2 - 4(1)(2) = 4 - 8 = -4 \). The negative discriminant indicates there are no real solutions.

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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 type of polynomial equation that can be written in the form \( ax^2 + bx + c = 0 \). These equations are characterized by the presence of the variable squared, or \( x^2 \). The main goal in solving a quadratic equation is to find the values of \( x \) (also known as roots or solutions) that satisfy the equation.
  • They can be solved using methods like factoring, completing the square, or the quadratic formula.
  • The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), provides an efficient way to find the solutions without needing to manipulate the original equation too much.
  • Understanding the discriminant \( b^2 - 4ac \) is crucial, as it indicates the nature of the equation's solutions.
Once established, solving these equations helps in analyzing fundamental relationships, such as projectile motion in physics or financial forecasting in econometrics. This exercise asks us to solve a quadratic equation derived from simplifying a more complex equation, emphasizing the universal applicability of such equations.
Fraction Simplification
Fraction simplification involves reducing a rational expression to its simplest form by finding a common denominator and combining like terms.
  • The common denominator is crucial when adding or subtracting fractions because it allows them to be combined into a single fraction. In this problem, the common denominator for \( \frac{1}{x-1} \) and \( \frac{2}{x^2} \) is \( x^2(x-1) \).
  • Once a common denominator is identified, each fraction must be rewritten so that they all share this denominator. This involves adjusting both the numerators and denominators accordingly.
  • Combining fractions over the common denominator then facilitates further simplification and solving for the variable \( x \).
This problem teaches the value of strategically simplifying expressions to tackle more complex equations effectively. The process of finding a common denominator and reconciling various components of equations is a staple skill across diverse areas of mathematics.
No Real Solutions
Some quadratic equations do not have roots that are real numbers. This occurs when the discriminant, \( b^2 - 4ac \), is negative.
  • The discriminant shows whether the roots are real or complex, with a negative value indicating the absence of real solutions.
  • For the equation \( x^2 - 2x + 2 = 0 \) we computed the discriminant to be \(-4\), which confirms there are no real roots.
  • Equations with negative discriminants result in complex numbers, and hence solutions go beyond the real number line.
Understanding when an equation has no real solutions helps in recognizing the limits of certain scenarios or predicting systems with no feasible outcomes. Knowing the nature of solutions that a quadratic equation can yield adds to the broader comprehension of mathematical problem-solving.

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Most popular questions from this chapter

Decimal Notation Write each number in decimal notation. (a) \(7.1 \times 10^{14}\) (b) \(6 \times 10^{12}\) (c) \(8.55 \times 10^{-3}\) (d) \(6.257 \times 10^{-10}\)

Gravity If an imaginary line segment is drawn between the centers of the earth and the moon, then the net gravitational force \(F\) acting on an object situated on this line segment is $$F=\frac{-K}{x^{2}}+\frac{0.012 K}{(239-x)^{2}}$$ where \(K>0\) is a constant and \(x\) is the distance of the object from the center of the earth, measured in thousands of miles. How far from the center of the earth is the "dead spot" where no net gravitational force acts upon the object? (Express your answer to the nearest thousand miles.) PICTURE CANT COPY

Complete the squares in the general equation \(x^{2}+a x+y^{2}+b y+c=0\) and simplify the result as much as possible. Under what conditions on the coefficients \(a, b,\) and \(c\) does this equation represent a circle? A single point? The empty set? In the case in which the equation does represent a circle, find its center and radius.

Manufacturer's Profit If a manufacturer sells \(x\) units of a certain product, revenue \(R\) and cost \(C\) (in dollars) are given by $$ \begin{array}{l} R=20 x \\ C=2000+8 x+0.0025 x^{2} \end{array} $$ Use the fact that profit \(=\) revenue \(-\) cost to determine how many units the manufacturer should sell to enjoy a profit of at least \(\$ 2400\).

Solving an Equation in Different Ways We have learned several different ways to solve an equation in this section. Some equations can be tackled by more than one method. For example, the equation \(x-\sqrt{x}-2=0\) is of quadratic type. We can solve it by letting \(\sqrt{x}=u\) and \(x=u^{2},\) and factoring. Or we could solve for \(\sqrt{x},\) square each side, and then solve the resulting quadratic equation. Solve the following equations using both methods indicated, and show that you get the same final answers. (a) \(x-\sqrt{x}-2=0 \quad\) quadratic type; solve for the radical, and square (b) \(\frac{12}{(x-3)^{2}}+\frac{10}{x-3}+1=0 \quad \begin{array}{l}\text { quadratic type; multiply } \\ \text { by } \mathrm{LCD}\end{array}\)
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