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Quadratic equations are a special type of polynomial equation. They have the general form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents an unknown variable. The defining feature of quadratic equations is that the highest degree (or the power) of \( x \) is 2.

These equations are fundamental in algebra and come up often because they describe parabola shapes when graphed on a coordinate plane.

You might encounter them in real-world scenarios, such as calculating the trajectory of a projectile or optimizing areas.

• When solving quadratic equations, there are multiple methods available: factoring, using the quadratic formula, and completing the square.
• Each method has its use, depending on the characteristics of the equation.

Completing the square, the method used in our given problem, is particularly useful for turning a quadratic into a perfect square trinomial. This makes it easier to find solutions.

A perfect square is a number that can be expressed as the product of an integer with itself. In the context of quadratic equations, a perfect square trinomial is an expression that takes the form \( (x - m)^2 \). Completing the square involves rewriting part of the quadratic as a perfect square trinomial.

The process of completing the square includes:

• Starting with the quadratic expression, rearrange it into the form \( ax^2 + bx \).
• Identify the coefficient of \( x \), then take half of this coefficient (let’s call it \( m \)), and square it to get \( m^2 \).
• Add and subtract \( m^2 \) inside the equation to form a complete square.

In our example, the quadratic \( x^2 - 2x \) was transformed into \( (x - 1)^2 \). The operation added \( 1^2 \) both inside and outside to balance the equation, helping to solve it further.

Solving an equation involves finding the value(s) of the unknown variables that make the equation true. For quadratic equations, once they are arranged in a solvable form (like a perfect square), we employ algebraic manipulations to arrive at solutions.

When a quadratic is set up as a perfect square trinomial, we can solve for \( x \) by taking the square root of both sides. Remember, squaring introduces both positive and negative solutions. This is why we consider both \( \sqrt{5} \) and \( -\sqrt{5} \) as part of solving the equation in our example.

• Taking square roots is a critical step that gives two possible solutions.
• After computing the square root, rearrange the equation by isolating \( x \), which may involve adding or subtracting terms on both sides.

For our quadratic, the solutions \( x = 1 + \sqrt{5} \) and \( x = 1 - \sqrt{5} \) were derived after employing the square root property and subsequent algebraic steps. These solutions satisfy the original equation.