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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 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 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.