Problem 5 - Algebra 2

Question

Solve each equation. Check each solution. $$ \frac{3}{x+1}=\frac{1}{x^{2}-1} $$

Step-by-Step Solution

Answer

The valid solution to the equation is \(x = 2\). Other solutions don't work when substituted back into the original equation.

Step 1: Cross-Multiplication

Start by cross-multiplying, turning the equation into: \(3(x^2 - 1) = 1(x + 1)\). This results in \(3x^2 - 3 = x + 1\)

Step 2: Simplify and Rearrange

Simplify by moving all terms to one side of the equation, thus forming a quadratic equation: \(3x^2 - x - 4 = 0\)

Step 3: Apply the Quadratic Formula

Apply the quadratic formula \(x = [-b ± sqrt(b^2 - 4ac)]/(2a)\): For the equation \(3x^2 - x - 4 = 0\), \(a = 3\), \(b = -1\), and \(c = -4\). Substituting these values into the quadratic formula yields \(x = \frac{1 ± sqrt(1 + 4(3)(4))}{2(3)}\)

Step 4: Simplify to find solutions

Simplify the expressions to find possible values of \(x\). This results in two potential solutions: \(x = \frac{1 + sqrt(49)}{6}\) or \(x = \frac{1 - sqrt(49)}{6}\), which simplify to \(x = 2\) and \(x = -2/3\)

Step 5: Check Solutions

Substitute \(x = 2\) and \(x = -2/3\) to the original equation to check if they are valid solutions. \(x = 2\) results into \(3/3=1/3\) and \(x = -2/3\) results into \(-3/3=-1\). The first solution is valid but the second one is not. So, only \(x = 2\) is the valid solution.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool in solving second-degree polynomial equations of the form \( ax^2 + bx + c = 0 \). When the quadratic equation is not easily factorable, the quadratic formula provides a method to find its roots. The formula is written as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \( a \), \( b \), and \( c \) are constants from the quadratic equation, and \( x \) represents the roots or solutions of the equation. Let's consider what each part of the formula does:

• -b: Flips the sign of \( b \), which shifts the parabola horizontally on the graph.
• \( b^2 - 4ac \): Known as the discriminant, determines the number and type of solutions.
  • If the discriminant is positive, there are two real and distinct solutions.
  • If it's zero, there is a single real and repeated solution.
  • If negative, the solutions are complex numbers.
• \( 2a \): Represents the denominator that scales down the result.

In our exercise, applying the quadratic formula, we found potential solutions: \( x = 2 \) and \( x = -\frac{2}{3} \). The formula handles all sorts of quadratic equations, making it a versatile approach.

Cross-Multiplication

Cross-multiplication is a simplification tool used in solving rational equations, where two fractions are set equal to each other. Consider the given equation \[ \frac{3}{x+1} = \frac{1}{x^2-1} \] To eliminate the fractions, we multiply both sides by the denominators of the opposite fraction, transforming the equation devoid of fraction.

• The process involves multiplying diagonally: \(3 \times (x^2 - 1)\) and \(1 \times (x + 1)\).
• This results in the equation: \(3(x^2 - 1) = (x + 1)\).

Cross-multiplication simplifies rational equations making them easier to solve. From here, you would simplify and reformulate into a more recognizable quadratic equation.

Checking Solutions

Checking solutions is a crucial step in any mathematical problem to confirm the validity of the obtained solutions. Once the possible solutions for the quadratic equation \( x = 2 \) and \( x = -\frac{2}{3} \) were found, each must be substituted back into the original equation to verify correctness.

• First, substitute \( x = 2 \) back into \( \frac{3}{x+1} = \frac{1}{x^2-1} \):
  • The left side becomes \( \frac{3}{3} = 1 \) and the right side becomes \( \frac{1}{3} \), showing this is a valid solution.
• Next, substitute \( x = -\frac{2}{3} \):
  • The left side is \( \frac{3}{\frac{1}{3}} = -9 \) while the right is negative but not equal, indicating it's not a valid solution.

Checking ensures that only true, meaningful solutions are considered. It's especially important in equations like these, where operations might result in extraneous solutions, highlighting the importance of thorough verification.