91Ó°ÊÓ

Quadratic equations are polynomial equations of the second degree. They have the general form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). In the given exercise, the equation \( (x-3)^{2}=36 \) is a quadratic equation because it can be expanded to fit that form. Here, we see that if we expand \((x-3)^2\), it becomes a quadratic expression:

• Start from \((x-3)^2 = x^2 - 6x + 9\)
• This equation suggests that the constant \( b = -6 \) and the linear term \( c = 9 \).

Indeed, quadratic equations are relations that generate a parabola when graphed. Solutions for these equations are often found where the parabola crosses the x-axis, but this is not always convenient for solving purposes. In the context of the original exercise, the Square Root Property offers a direct path to simplify and solve.

Algebraic solutions refer to the process of solving equations using algebraic operations and methods like factoring or applying properties. In our scenario, the Square Root Property is the key algebraic tool used. The Square Root Property states:

• If an equation is in the form \( a^2 = b \), it implies \( a = \sqrt{b} \) or \( a = -\sqrt{b} \).
• This property allows us to isolate the variable by considering both the positive and negative square roots of a number.

For the given problem, we have \((x-3)^2 = 36\). Here, the algebraic solution involves applying the property to obtain equations:

• First, \( (x-3) = \sqrt{36} \), which gives \( x = 9 \).
• Second, \( (x-3) = -\sqrt{36} \), which gives \( x = -3 \).

The algebraic solution confirms both positive and negative roots, offering both possible values of \( x \) that satisfy the equation.

Solving equations is a fundamental aspect of algebra, involving finding values of variables that make the equations true. In this exercise, the aim is to solve a quadratic equation using the Square Root Property. Let's break down the steps:

• Identify the equation: Here, we've \((x-3)^2 = 36\). Recognize it's set for the Square Root Property.
• Apply the property: Recognize that \( (x-3) \) equals either \( \sqrt{36} \) or \( -\sqrt{36} \), breaking it into two solvable equations.
• Solve for \( x \):
  • For \( (x-3) = \sqrt{36} \), solve to get \( x = 9 \).
  • For \( (x-3) = -\sqrt{36} \), solve to get \( x = -3 \).

These steps simplify solving equations that involve squares to their roots, streamlining the process and ensuring you correctly address both root solutions. This method is part of the broader strategy used in algebra to solve different types of equations, assuring the equations are both simplified and accurately balanced.