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The factoring method for solving quadratic equations is often a favored approach because of its simplicity and speed. It involves expressing the quadratic equation in the form \[ (x - p)(x - q) = 0 \]where \( p \) and \( q \) are numbers that satisfy both of the following conditions:

• They multiply to give you the constant term \( c \) at the end of the equation.
• They add up to give you the middle coefficient \( b \).

Once you have the equation written this way, finding the solutions becomes straightforward because you can set each factor equal to zero. For instance, with an equation like \[ x^2 - 5x + 6 = 0 \]we look for two numbers that multiply to 6 and add up to -5. The numbers \(-2\) and \(-3\) work since \((-2) \times (-3) = 6\) and \(-2) + (-3) = -5\). You then solve the derived linear equations:\( x - 2 = 0 \) which gives \( x = 2 \), and \( x - 3 = 0 \) which gives \( x = 3 \).
The solutions \( x = 2 \) and \( x = 3 \) are the roots of the quadratic equation, meaning they are the values where the equation equals zero.

Completing the square is a systematic way to solve a quadratic equation by turning it into a perfect square trinomial. The goal is to manipulate the original equation to a form like:\[ (x - d)^2 = e \]Let's see this in action with our example equation \( x^2 - 5x + 6 = 0 \).

• First, move the constant term to the right: \( x^2 - 5x = -6 \).
• To complete the square, add the square of half the coefficient of \( x \), which in this case is \( \left(\frac{-5}{2}\right)^2 = \frac{25}{4} \), to both sides.
• You now have \( x^2 - 5x + \frac{25}{4} = \frac{1}{4} \). The left side is a perfect square trinomial \((x - \frac{5}{2})^2 \).

From \( (x - \frac{5}{2})^2 = \frac{1}{4} \), solve for \( x \) using the property \( a^2 = b\) which gives \( a = \pm \sqrt{b} \). So you get:

• \( x - \frac{5}{2} = \pm \frac{1}{2} \)
• which yields two solutions: \( x = 3 \) and \( x = 2 \).

This method is particularly useful when the quadratic cannot be easily factored, and it forms the basis for deriving the quadratic formula.

The quadratic formula is a foolproof method for solving any quadratic equation \[ ax^2 + bx + c = 0 \]It is derived by completing the square in the general case and provides the solutions directly using the formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In this application, identify the coefficients \( a \), \( b \), and \( c \) from your equation. Let's look at \( x^2 - 5x + 6 = 0 \)where \( a = 1 \), \( b = -5 \), and \( c = 6 \). Substitute these values into the formula:

• Calculate the discriminant \( b^2 - 4ac \), which informs the nature of the roots. Here, it's \( 25 - 24 = 1 \).
• Plug everything into the formula: \[ x = \frac{-(-5) \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2} \]

This simplifies to two possible solutions: \( x = \frac{6}{2} = 3 \) and \( x = \frac{4}{2} = 2 \).
The quadratic formula is extremely powerful, offering a solution pathway regardless of complex numbers or irrational roots. It guarantees solving any quadratic equation when other methods may be tedious or impossible.