91Ó°ÊÓ

Factoring is a key technique to solve quadratic equations. When you factor, you break down an equation into simpler components that are easier to solve. In this exercise, the equation (2x + 7)(x^2 + 2x - 3) = 0 is already partially factored. Your task is to solve for the variables in these factors. This typically means setting each factor equal to zero and solving independently. For example, to factor x^2 + 2x - 3, you can look for two numbers that multiply to -3 (the constant term) and add to 2 (the middle term coefficient). Here, those numbers are 3 and -1, leading to the factorization: (x + 3)(x - 1). This process of breaking down equations can make it less overwhelming and manageable.

When direct factoring is challenging or impossible, the quadratic formula is a reliable alternative. The quadratic formula is given by: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] where a, b, and c are the coefficients from the equation ax^2 + bx + c = 0. This formula helps us find the roots of any quadratic equation. For example, in the equation x^2 + 2x - 3 = 0, the coefficients are a = 1, b = 2, and c = -3. Plugging these values into the quadratic formula yields the solutions: \[ x = \frac{{-2 \pm \sqrt{{4 + 12}}}}{2} = \frac{{-2 \pm 4}}{2} \] resulting in x = 1 and x = -3. Even if factoring is not straightforward, this method always provides a way to find the solutions.

After solving the quadratic equation, it is important to check your solutions to confirm they are correct. This involves substituting each solution back into the original equation and verifying that it holds true. For example, consider the equation: \( (2x + 7)(x^2 + 2x - 3) = 0 \). The solutions are x = -\frac{7}{2}, x = -3, and x = 1. To check x = -\frac{7}{2}, substitute this value into the equation: \[ (2(-\frac{7}{2}) + 7)((-\frac{7}{2})^2 + 2(-\frac{7}{2}) - 3) = 0 \] When simplified correctly, it verifies that the left side equals 0, confirming x = -\frac{7}{2} is a solution. Repeat this for x = -3 and x = 1. This step ensures no mistakes are missed and the roots truly satisfy the original equation.