Problem 11 - College Algebra with Modeling and Visualization

Question

Find all real solutions. Check your results. $$ \frac{1}{x+2}+\frac{1}{x}=1 $$

Step-by-Step Solution

Answer

The real solutions are \\(x = \sqrt{2}\\) and \\(x = -\sqrt{2}\\).

Step 1: Eliminate Fractions

To eliminate the fractions, find a common denominator, which is \(x(x+2)\). Multiply every term in the equation by this common denominator:\[ x(x+2) \left( \frac{1}{x+2} + \frac{1}{x} \right) = x(x+2) \cdot 1 \]This yields:\[ x + (x+2) = x^2 + 2x \]

Step 2: Simplify the Equation

Now simplify the expression obtained:\[ x + x + 2 = x^2 + 2x \]Combine like terms:\[ 2x + 2 = x^2 + 2x \]

Step 3: Rearrange into a Quadratic Equation

Move all terms to one side of the equation so that it equals zero:\[ 0 = x^2 + 2x - 2x - 2 \]This simplifies to:\[ 0 = x^2 - 2 \]Thus, the equation becomes a quadratic:\[ x^2 - 2 = 0 \]

Step 4: Solve the Quadratic Equation

To solve \(x^2 - 2 = 0\), add 2 to both sides:\[ x^2 = 2 \]Take the square root of both sides:\[ x = \pm\sqrt{2} \]

Step 5: Verify Solutions

Substitute \(x = \sqrt{2}\) and \(x = -\sqrt{2}\) back into the original equation to verify:- For \(x = \sqrt{2}\), the equation becomes \(\frac{1}{\sqrt{2}+2} + \frac{1}{\sqrt{2}} \), which upon adding gives 1.- For \(x = -\sqrt{2}\), the equation becomes \(\frac{1}{2-\sqrt{2}} + \frac{1}{-\sqrt{2}} \), which also adds to 1.Both values of \(x\) satisfy the original equation.

Key Concepts

Real Solutions

In the realm of algebra, when we solve equations, we often look for 'real solutions'. Real solutions refer to the values of the variable that satisfy the equation and can be found on the real number line. These solutions do not involve imaginary numbers. To check if a value is a real solution:

• Substitute the value back into the original equation.
• Perform the necessary arithmetic operations.
• Ensure that the equality holds true.

For example, the values \(x = \pm \sqrt{2}\) are real solutions to the quadratic equation obtained from our problem. After substituting these values back, the equation balances, confirming their validity.

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. It takes on the general form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). The solution of a quadratic equation can be found using several methods, such as factoring, completing the square, or the quadratic formula. For the problem above, we derived \(x^2 - 2 = 0\). This is a simple form of a quadratic equation where:

• The coefficient \(a = 1\)
• The coefficient \(b = 0\) is implied
• The constant \(c = -2\)

The solutions to this equation can be obtained by adding 2 to both sides and then taking the square root, leading us to the solutions \(x = \pm \sqrt{2}\). Quadratic equations often have two potential solutions, as seen here.

Common Denominator

A common denominator is a shared multiple of the denominators of two or more fractions. Finding a common denominator is crucial when trying to combine or compare fractions. For instance, when dealing with an equation that involves fractions, such as our original exercise, using a common denominator can help eliminate the fractions and simplify the equation. Here's how it works:

• Identify the least common multiple of the denominators involved.
• Multiply every term of the equation by this common denominator.
• Simplify the resulting equation.

In our exercise, the denominators of the fractions were \(x+2\) and \(x\). The common denominator is \(x(x+2)\). Multiplying through by this helps in removing the fractions, making further algebraic manipulation more simple and straightforward. This step is crucial for simplifying complex equations into solvable forms, particularly when dealing with quadratic equations.