Problem 37 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \left(\frac{1}{x+1}\right)^{2}-2\left(\frac{1}{x+1}\right)-8=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = -\frac{3}{4} \) and \( x = -\frac{3}{2} \).

Step 1: Introduce a Substitution

Notice that the expression \( \left(\frac{1}{x+1}\right) \) appears repeatedly. Let's introduce a substitution to make the equation simpler. Let \( y = \frac{1}{x+1} \). The equation now becomes \( y^2 - 2y - 8 = 0 \).

Step 2: Solve the Quadratic Equation

We now solve the quadratic equation \( y^2 - 2y - 8 = 0 \). Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \) with \( a = 1 \), \( b = -2 \), and \( c = -8 \). Calculating, we have \( y = \frac{2 \pm \sqrt{4 + 32}}{2} \), leading to \( y = \frac{2 \pm 6}{2} \).

Step 3: Calculate the Roots for y

Solving \( y = \frac{2 + 6}{2} = 4 \) and \( y = \frac{2 - 6}{2} = -2 \), we find the values \( y = 4 \) and \( y = -2 \).

Step 4: Reverse the Substitution

Remembering the substitution \( y = \frac{1}{x+1} \), we now reverse it: \( \frac{1}{x+1} = 4 \) and \( \frac{1}{x+1} = -2 \).

Step 5: Solve for x

For \( \frac{1}{x+1} = 4 \), we have \( x+1 = \frac{1}{4} \), leading to \( x = -\frac{3}{4} \). For \( \frac{1}{x+1} = -2 \), we have \( x+1 = -\frac{1}{2} \), leading to \( x = -\frac{3}{2} \).

Step 6: Verify Solutions

Plugging \( x = -\frac{3}{4} \) and \( x = -\frac{3}{2} \) back into the original equation verifies they are correct as they satisfy \( \left(\frac{1}{x+1}\right)^{2} - 2\left(\frac{1}{x+1}\right) - 8 = 0 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of a quadratic equation, which is generally expressed as \( ax^2 + bx + c = 0 \). The formula itself is: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula works by calculating the values of \( y \) that satisfy the equation based on the coefficients \( a \), \( b \), and \( c \). It is useful because it can provide both real and complex solutions.

• \( -b \): This term starts with the opposite of \( b \), providing symmetry around the axis of the parabola formed by the quadratic equation.
• \( \pm \sqrt{b^2 - 4ac} \): This part determines whether the solutions are real or complex, known as the discriminant. If it's positive, you have real solutions.
• \( 2a \): This denominator helps to scale the solution correctly depending on the horizontal stretch of the parabola.

In our original equation, once we substituted \( y = \frac{1}{x+1} \), it became simple to apply the quadratic formula. By solving \( y^2 - 2y - 8 = 0 \) with \( a = 1 \), \( b = -2 \), and \( c = -8 \), we derived the solutions, \( y = 4 \) and \( y = -2 \). These roots can then be re-substituted to find \( x \).

Substitution Method

The substitution method is an effective technique used to simplify complex equations, making them easier to solve. It involves replacing part of the equation with a new variable. In the original exercise, the fraction \( \left( \frac{1}{x+1} \right) \) appeared multiple times, indicating a candidate for substitution. By setting \( y = \frac{1}{x+1} \), it transformed our complicated equation into the much simpler quadratic form \( y^2 - 2y - 8 = 0 \). Here are the benefits of using the substitution method:

• Makes complex expressions simpler: It reduces the equation to a well-known form like quadratic, which is easier to handle.
• Avoids cumbersome calculations: By lowering the cognitive load, it allows focusing on solving rather than deciphering the equation.

Once the simpler equation is solved, you can substitute back to find the original variable. This process highlights the power of maintaining mathematical clarity and organization.

Real Solutions

Real solutions are those that can be plotted on a number line, and they occur when the discriminant in the quadratic formula is zero or positive. For the given equation, the discriminant was calculated as 36 (from the expression \( 4 + 32 \)), which is positive, indicating real solutions. To understand why a positive discriminant results in real solutions:

• A positive discriminant means the expression \( \sqrt{b^2 - 4ac} \) has a real number solution, resulting in two distinct roots.
• When \( b^2 - 4ac = 0 \), there's exactly one real root, referred to as a repeated or double root.
• If \( b^2 - 4ac \) were negative, it would imply imaginary solutions, inaccessible on a real number line.

Hence, the quadratic \( y^2 - 2y - 8 = 0 \) revealed two real roots, \( y = 4 \) and \( y = -2 \), which relate back to \( x \) through our initial substitution, leading to the solutions \( x = -\frac{3}{4} \) and \( x = -\frac{3}{2} \). Understanding how to interpret discriminant results provides essential insights into the nature of the solutions you may encounter.