Problem 24 - College Algebra

Question

Solve the quadratic equation by using the square root property. $$ (x-5)^{2}=4 $$

Step-by-Step Solution

Answer

The solutions are \(x = 3\) and \(x = 7\).

Step 1: Understanding the Equation

The given equation is \((x-5)^2 = 4\). This equation is in the form \((a)^2 = b\), where \(a = x - 5\) and \(b = 4\).

Step 2: Apply the Square Root Property

The square root property states that if \((a)^2 = b\), then \(a = \pm \sqrt{b}\). Apply this property to the equation: \(x - 5 = \pm \sqrt{4}\).

Step 3: Simplifying the Equation

Calculate the square root of 4. Since \(\sqrt{4} = 2\), substitute into the equation: \(x - 5 = \pm 2\). This gives us two separate equations to solve: \(x - 5 = 2\) and \(x - 5 = -2\).

Step 4: Solve for x, Case 1

For the first equation \(x - 5 = 2\), add 5 to both sides to solve for \(x\):\[x - 5 + 5 = 2 + 5\] \[x = 7\]

Step 5: Solve for x, Case 2

For the second equation \(x - 5 = -2\), add 5 to both sides to solve for \(x\):\[x - 5 + 5 = -2 + 5\] \[x = 3\]

Step 6: Write the Solution Set

The solutions from the two cases are \(x = 7\) and \(x = 3\). Thus, the solution set is \(\{3, 7\}\).

Key Concepts

Square Root Property

The square root property is a valuable tool in solving quadratic equations, especially those of the form \(a^2 = b\). In our given equation \( (x - 5)^2 = 4\), it fits this form where \(a = x - 5\) and \(b = 4\). The square root property tells us that if \(a^2 = b\), then \(a = \pm \sqrt{b}\). This means that when you have a perfect square set equal to a number, you can take the square root of both sides, which will always give you two possible equations due to the positive and negative solutions. It's important to remember this when solving such problems, as it helps us find both possible values for \(a\). In essence, the square root property expands our solution set by considering both the positive and negative roots from the square root operation.

Solving Equations

Solving equations often requires transforming the equation into a form that is easier to work with. This includes both isolating terms and simplifying expressions. When using the square root property, once we have the simpler form of \(a = \pm \sqrt{b}\), we can split this into two separate linear equations to solve.
For instance, with our equation \(x - 5 = \pm \sqrt{4}\), it breaks down into \(x - 5 = 2\) and \(x - 5 = -2\). Each equation is handled independently by algebraic manipulation to solve for \(x\).

• Add 5 to both sides to isolate \(x\).
• Evaluate each solution separately to find distinct values.

This systematic approach ensures you don’t miss any potential solutions.

Solution Set

The solution set is the collection of all possible values of \(x\) that satisfy the equation. After solving \(x - 5 = 2\) and \(x - 5 = -2\), you find \(x = 7\) and \(x = 3\), respectively. Together, these form the solution set for the quadratic equation we started with, which is \{3, 7\}.
A solution set shows every possible solution to the given problem, acknowledging that more than one value can satisfy the equation. Maintaining a correct solution set is crucial as it verifies the completeness of your work in solving quadratic equations. Always check each solution by plugging it back into the original equation to ensure accuracy.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to isolate the variable of interest. In our equation, \(x - 5 = \pm 2\), manipulation involves a few key steps:

• Adding or subtracting terms from both sides of an equation to maintain balance and isolate \(x\).
• Simplifying each equation to find the simplest form of \(x\).

For example, by adding 5 to both sides of each split equation, you systematically reduce the equation to a single x-term. This process is essential in solving not just quadratic equations, but any algebraic equations in general, ensuring clarity and accuracy in finding the correct solutions.