Problem 22
Question
Use the even-root property to solve each equation. $$(a-2)^{2}=8$$
Step-by-Step Solution
Verified Answer
\(a = 2 \text{±} 2√2\).
1Step 1: Understand the Even-Root Property
The even-root property states that if \( x^2 = k \), then \( x = \boxed{\text{±}\text{\text{s}}\text{r}t{k}}\). In this case, \((a-2)^2 = 8\).
2Step 2: Apply the Square Root
Take the square root of both sides of the equation: \(\text{\text{s}}\text{r}t{(a-2)^2} = \text{\text{±}\text{s}}\text{rt}{8}\). This simplifies to:\(|a-2| = \boxed{\text{2√2}}\).
3Step 3: Separate into Two Equations
Since \(|a-2| = 2√2\), we can create two separate equations: \(a-2 = 2√2\) and\(a-2 = −2√2\).
4Step 4: Solve for a - Equation 1
For \(a-2 = 2√2\):\(a = 2 + 2√2\).
5Step 5: Solve for a - Equation 2
For \(a-2 = -2√2\):\(a = 2 - 2√2\).
Key Concepts
Solving EquationsSquare RootAbsolute Value
Solving Equations
Solving equations is like solving puzzles. We have an unknown value, often represented by a letter like \( x \) or \( a \), and our job is to figure out what that value is. Equations can represent real-world problems and mathematical relationships. Here are some steps to solve equations:
- First, isolate the variable on one side of the equation. This allows us to find its value.
- Use mathematical operations like addition, subtraction, multiplication, and division to simplify.
- Always check your solutions by substituting them back into the original equation.
Square Root
The square root operation is the opposite of squaring a number. If you square 3, you get 9, and if you take the square root of 9, you get 3. Mathematically, this relationship is shown as: \( \text{if } x^2 = k, \text{ then } x = \boxed{\text{±}\text{sqrt}(k)} \).
In the exercise, we applied the square root to both sides of the equation \( (a-2)^2 = 8 \). Taking the square root of both sides, we get: \[ \text{sqrt} ((a-2)^2) = \text{±}\text{sqrt}(8) \]. This simplifies to: \[ |a-2| = 2\text{sqrt}(2) \]. This step is key because it makes the equation easier to solve. Keep in mind that taking the square root introduces both positive and negative solutions. These two potential values help us find all possible solutions to the original problem.
In the exercise, we applied the square root to both sides of the equation \( (a-2)^2 = 8 \). Taking the square root of both sides, we get: \[ \text{sqrt} ((a-2)^2) = \text{±}\text{sqrt}(8) \]. This simplifies to: \[ |a-2| = 2\text{sqrt}(2) \]. This step is key because it makes the equation easier to solve. Keep in mind that taking the square root introduces both positive and negative solutions. These two potential values help us find all possible solutions to the original problem.
Absolute Value
The absolute value of a number is its distance from zero on the number line, regardless of direction. It is always non-negative. For example, \( |3| = 3 \) and \( |-3| = 3 \). Absolute value is represented by vertical bars around the number or expression, like |x|.
In solving the exercise, we reached \( |a-2| = 2\text{sqrt}(2) \). This implies two separate equations:
In solving the exercise, we reached \( |a-2| = 2\text{sqrt}(2) \). This implies two separate equations:
- \( a-2 = 2\text{sqrt}(2) \)
- \( a-2 = -2\text{sqrt}(2) \)
- For \( a-2 = 2\text{sqrt}(2) \), we add 2 to both sides, resulting in \( a = 2 + 2\text{sqrt}(2) \).
- For \( a-2 = -2\text{sqrt}(2) \), we add 2 to both sides, resulting in \( a = 2 - 2\text{sqrt}(2) \).
Other exercises in this chapter
Problem 22
Graph each quadratic function, and state its domain and range. $$y=\frac{1}{3} x^{2}-6$$
View solution Problem 22
Solve each equation by using the quadratic formula. $$p^{2}+6 p+4=0$$
View solution Problem 23
Graph each quadratic function, and state its domain and range. $$f(x)=-2 x^{2}+5$$
View solution Problem 23
Solve each equation by using the quadratic formula. $$-x^{2}-5 x+1=0$$
View solution