91Ó°ÊÓ

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants with \(a eq 0\). Solving these equations involves finding the values of \(x\) that satisfy the equation.
When working with quadratic equations, there are several methods to find the solution:

• Factoring: Splitting the equation into two factors that multiply to zero.
• Quadratic Formula: Using \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the Square: Rewriting the equation in the form \((x + p)^2 = q\).
• Graphing: Plotting the equation and finding where it intersects the x-axis.

In this exercise, the chosen method is completing the square. This method is particularly useful when the equation cannot be easily factored, offering a precise way to find the roots of the quadratic equation.

A perfect square trinomial is a quadratic expression that can be expressed as the square of a binomial. In other words, it's an expression of the form \((x + a)^2\) or \((x - a)^2\), which expands to \(x^2 + 2ax + a^2\).
The process of completing the square involves converting a quadratic equation into a perfect square trinomial. By doing so, the expression becomes much easier to solve. Given the equation \(s^2 + 2s = 6\), the term we need to add to make a perfect square trinomial is \((\frac{2}{2})^2 = 1\).
After adding \(1\) to both sides, we get \((s + 1)^2 = 7\). The left side is now a perfect square trinomial, simplifying the path to solving the equation. Perfect square trinomials streamline the solving process by creating a direct equation to work with.

Once you have turned a quadratic equation into the form of a perfect square, the Square Root Method becomes quite handy. This approach is about taking the square root of both sides of the equation to find the value of the variable.
In our previous steps, the equation \((s + 1)^2 = 7\) was formed. To solve for \(s\), apply the square root method:

• Take the square root of both sides: \(s + 1 = \pm \sqrt{7}\).
• Isolate \(s\) by subtracting \(1\) from both sides: \(s = -1 \pm \sqrt{7}\).

The result \(s = -1 + \sqrt{7}\) and \(s = -1 - \sqrt{7}\) provides the two possible solutions for the equation. Remember, when taking square roots, consider both the positive and negative roots to account for all potential solutions.