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A perfect square trinomial is a special type of algebraic expression. It's the result of squaring a binomial, which simply means multiplying a binomial by itself. For example,

• a binomial like \((x + a)^2\)
• results in the trinomial \(x^2 + 2ax + a^2\).

In the context of the exercise, we identified \(x^2 + 18x + 81\) as a perfect square trinomial. This is because it can be rewritten as \((x + 9)^2\).Here's why:

• The first term, \(x^2\), remains the same.
• The middle term, \(18x\), equals \(2 \times x \times 9\), fitting the pattern of \(2ax\).
• The last term, \(81\), is \(9^2\), fitting the pattern of \(a^2\).

Identifying perfect square trinomials allows one to rewrite them in a simpler form, like \((x + 9)^2\), which is much easier to solve.

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\). The degree of the polynomial is 2, which means it is characterized by the highest exponent of \(x\), namely \(x^2\). Such equations can have two, one, or no solutions, depending on the values of \(a\), \(b\), and \(c\).Quadratic equations can be solved using various methods, including:

• factoring,
• using the quadratic formula,
• completing the square,
• and applying the Square Root Property.

In this specific instance, recognizing the perfect square trinomial allowed us to employ the Square Root Property effectively. Once rewritten, it simplifies the process of finding the roots or solutions of the equation. In many cases, simplifying to a binomial square can turn a seemingly complex equation into a direct calculation.

Square roots are mathematical quantities that are multiplied by themselves to yield a given number. They are often used to solve equations where a variable is squared.
For example, \(\sqrt{25} = 5\) because \(5 \times 5 = 25\).Similarly, \(\sqrt{25} = -5\) since \(-5 \times -5 = 25\).This dual solution is why we say \(\pm\sqrt{b}\).
The Square Root Property is useful in equations like \((x + 9)^2 = 25\). By taking the square root of both sides, it simplifies:

• On the left, you get \((x + 9)\).
• On the right, you have \(\pm 5\).

With these values, the equation becomes two simpler linear equations: \(x + 9 = 5\) and \(x + 9 = -5\). Solving these equations yields the values of \(x\), demonstrating how square roots simplify solving equations involving squares.