- The original equation is \((x-7)^2 = 36\).
- Applying the square root property yields \(x - 7 = 6\) or \(x - 7 = -6\).
- Solving these linear equations gives \(x = 13\) and \(x = 1\).
- The correct options are \(\boxed{b, c}\).
Explanation
- Analyzing the Quadratic Equation
The goal is to solve the equation \((x-7)^2 = 36\) for \(x\). This is a quadratic equation where the expression \((x-7)\) is squared. To solve for \(x\), we need to isolate the variable. The most straightforward approach is to apply the square root property, which states that if \(u^2 = k\), then \(u = \sqrt{k}\) or \(u = -\sqrt{k}\). In this case, \(u = x-7\) and \(k = 36\).
- Solving the Equations
Applying the square root property, we take the square root of both sides:
\[\sqrt{(x-7)^2} = \pm\sqrt{36}\]
Since \(\sqrt{36} = 6\), the equation simplifies to:
\[x - 7 = \pm 6\]
This gives us two separate linear equations to solve for \(x\):
- \(x - 7 = 6\)
- \(x - 7 = -6\)
- Identifying the Correct Answers
Now, we solve each linear equation individually:
- For \(x - 7 = 6\), add 7 to both sides:
\[x = 6 + 7 = 13\]
- For \(x - 7 = -6\), add 7 to both sides:
\[x = -6 + 7 = 1\]
Thus, the two solutions for the equation are \(x = 13\) and \(x = 1\). Comparing these results with the given options, we find that \(x = 13\) (option b) and \(x = 1\) (option c) are the correct choices.
Examples
Understanding how to solve square equations is fundamental in various real-life scenarios, such as architectural design or physics simulations. For instance, if you are designing a square courtyard with a specific area, you use square roots to determine the length of each side. If an area is \(36 \text{ m}^2\), you know the side length is \(\sqrt{36} = 6 \text{ m}\). Mastering this allows you to reverse such calculations to determine unknown dimensions or rates in many technical fields.