Problem 25 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \frac{1}{x-1}-\frac{2}{x^{2}}=0 $$

Step-by-Step Solution

Answer

There are no real solutions.

Step 1: Understand the Equation

We are given the equation \( \frac{1}{x-1} - \frac{2}{x^2} = 0 \). This is a rational equation involving fractions with variables in the denominator. The goal is to find values of \( x \) that satisfy this equation.

Step 2: Eliminate Fractions

To solve the equation, eliminate the fractions by finding a common denominator. The common denominator for \( \frac{1}{x-1} \) and \( \frac{2}{x^2} \) is \( x^2(x-1) \). Multiply every term by this common denominator to get rid of the fractions:\( x^2(x-1) \cdot \frac{1}{x-1} - x^2(x-1) \cdot \frac{2}{x^2} = 0 \).

Step 3: Simplify the Equation

After multiplying through by the common denominator, simplify:1. First term: \( x^2(x-1) \cdot \frac{1}{x-1} = x^2 \).2. Second term: \( x^2(x-1) \cdot \frac{2}{x^2} = 2(x-1) \).This simplifies the equation to: \( x^2 - 2(x-1) = 0 \).

Step 4: Expand and Rearrange

Expand the equation and rearrange terms:\( x^2 - 2x + 2 = 0 \).

Step 5: Solve the Quadratic Equation

Now solve the quadratic equation \( x^2 - 2x + 2 = 0 \) using the quadratic formula where \( a = 1 \), \( b = -2 \), and \( c = 2 \):\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).Substitute the values:\( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(2)}}{2(1)} \).

Step 6: Calculate the Discriminant

Calculate the discriminant \( \Delta = b^2 - 4ac \):\( \Delta = (-2)^2 - 4 \cdot 1 \cdot 2 = 4 - 8 = -4 \).Since the discriminant is negative, there are no real solutions to the equation.

Key Concepts

Common Denominators

When solving rational equations, identifying a common denominator is crucial. This allows you to eliminate fractions, simplifying the equation considerably. In our exercise, we had the expression \( \frac{1}{x-1} - \frac{2}{x^2} = 0 \). Here, the expressions have different denominators: \( x-1 \) and \( x^2 \). To simplify these terms into one equation, we need a common denominator.

To find this, look for the least common multiple of the different denominators. The least common multiple of \( x-1 \) and \( x^2 \) is \( x^2(x-1) \). Once determined, you multiply each term by this common denominator. This action helps eliminate the fractions:

• Multiplying \( x^2(x-1) \) with \( \frac{1}{x-1} \) results in \( x^2 \).
• Doing the same with \( \frac{2}{x^2} \) results in \( 2(x-1) \).

By removing fractions, the task converts into simpler algebraic expressions, making it easier to solve within fewer steps.

Quadratic Equations

Many algebraic problems reduce to quadratic equations, taking on the form \( ax^2 + bx + c = 0 \). Our particular equation simplifies to \( x^2 - 2x + 2 = 0 \) after eliminating fractions and simplifying.

To solve any quadratic equation, there are several methods available:

• Factoring: Works well if the equation can be expressed as a product of binomials.
• Completing the Square: Involves rearranging the equation to create a perfect square trinomial.
• Quadratic Formula: A universal method where solutions can be found using \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In our exercise, we used the quadratic formula because factoring isn't straightforward for \( x^2 - 2x + 2 \), allowing us to find whether solutions are real or complex.

Discriminants

The discriminant is a key component in the quadratic formula, given by \( b^2 - 4ac \). It helps determine the nature of the roots for a quadratic equation. In the equation \( x^2 - 2x + 2 = 0 \), the coefficients are \( a = 1 \), \( b = -2 \), and \( c = 2 \).

The discriminant tells us:

• Positive Discriminant: Two real and distinct solutions.
• Zero Discriminant: One real, repeated solution (the vertex of the parabola touches the x-axis).
• Negative Discriminant: No real solutions, only complex ones (parabola does not intersect the x-axis).

In our exercise, the discriminant \( (-2)^2 - 4(1)(2) \) evaluates to \(-4\). This negative value indicates there are no real solutions, aligning with the fact that the graph of the equation does not intersect the x-axis.