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A quadratic equation is a second-degree polynomial equation in the form of \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). It typically has the following characteristics:

• The highest power of the variable, \( x \), is 2.
• It can have up to two real solutions, which are the values of \( x \) that satisfy the equation.

In the original exercise, our quadratic equation is \( x^2 - 10x = 1 \). To solve it, we aim to isolate the quadratic and linear terms from the constant to work towards completing the square. This involves transforming the equation such that we can easily apply other solution techniques, like using the square root property.

A perfect square trinomial is a special form of polynomial. It can be written as \( (x + d)^2 = x^2 + 2dx + d^2 \), where \( d \) is any real number. This form makes equations easier to solve, especially when solving by completing the square.In the context of our exercise, we began with \( x^2 - 10x \). By completing the square, we modified it to become a perfect square trinomial. Here's how:

• Took half of the coefficient of \( x \) (which was -10), resulting in -5.
• Squared this half: \((-5)^2\), resulting in 25.
• Added and subtracted this 25 to the equation.

By adding 25, the expression \( x^2 - 10x + 25 \) now transformed into the perfect square trinomial \((x - 5)^2\). This modification allows for simplifying the equation and is foundational when solving equations via the square root property.

The square root property is a method that allows us to solve equations once they are in the form of a squared term set equal to a constant. For any equation in the form \( (x - d)^2 = k \), this property allows us to directly take the square root of both sides:

• \( x - d = \pm \sqrt{k} \)

In the example from the exercise, after forming the perfect square \((x - 5)^2 = 26\), we applied the square root property:

• \( x - 5 = \pm \sqrt{26} \)

The "\( \pm \)" symbol indicates that there are two potential solutions — a crucial aspect when dealing with quadratic equations. Finally, to find \( x \), we moved the constant term by adding 5 to each side, resulting in the solutions \( x = 5 + \sqrt{26} \) and \( x = 5 - \sqrt{26} \). This approach is not only simple but also very effective for such quadratic expressions.