Problem 24 - Algebra A Combined Function

Question

Use the square root property to solve each equation. $$ (x+10)^{2}=11 $$

Step-by-Step Solution

Answer

The solutions are \(x = -10 + \sqrt{11}\) and \(x = -10 - \sqrt{11}\).

Step 1: Identify Equation Form

The given equation is \((x + 10)^2 = 11\), which is already in the form suitable for applying the square root property.

Step 2: Apply Square Root Property

The square root property states that if \(a^2 = b\), then \(a = \pm \sqrt{b}\). Apply this to obtain:\[x + 10 = \pm \sqrt{11}\]

Step 3: Solve for x

Isolate \(x\) by subtracting 10 from both sides. This gives:\[x = -10 \pm \sqrt{11}\]This results in two possible solutions:\[x_1 = -10 + \sqrt{11} \quad \text{and} \quad x_2 = -10 - \sqrt{11}\]

Step 4: Simplify Solutions

Since \(\sqrt{11}\) is already simplified, the solutions are:\[x = -10 + \sqrt{11}\quad \text{and} \quad x = -10 - \sqrt{11}\]

Key Concepts

Solving Equations

When we talk about solving equations, our goal is to find all the possible values that can be substituted for the variable to make the equation true. It's like figuring out a puzzle where the unknown value needs to be found.
The types of operations and techniques used to solve an equation depend on its structure and form.
In the context of simple algebraic equations, you might just need to use basic arithmetic operations like addition, subtraction, multiplication, or division. However, for more complex equations, different methods might be applied, such as factoring, using the quadratic formula, or in our case, employing the square root property.

• Firstly, identify the form of the equation to see what technique is suitable.
• Secondly, apply the correct mathematical operation or property, such as the square root property.
• Finally, isolate the variable to find the solution(s).

The step-by-step approach helps in breaking down the problem and applying the right strategy to get the correct answers.

Quadratic Equations

Quadratic equations are a special category of equations that have a variable raised to the power of 2, or squared. They are typically in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is the variable.
Quadratic equations can have two solutions, one solution, or no real solutions, depending on the discriminant, which is part of the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). However, not all quadratic equations require this complex formula.
In some cases, like when we have \((x+10)^2=11\), the equation is already in a format where we can directly apply the square root property to solve it.

• Such form makes it easier because it depicts a perfect square that can be easily "un-squared" with a proper operation.
• The focus here becomes taking the square root of both sides while remembering to include both the positive and negative roots.
• This method simplifies the process especially when the equation is designed as a perfect square.

Understanding the structure of quadratic equations allows us to choose the easiest and quickest way to find solutions.

Mathematical Operations

In solving any equation, mathematical operations are fundamental tools that help manipulate and simplify expressions.
The core operations include addition, subtraction, multiplication, and division. When solving equations, these operations help to isolate the variable, making it easier to find the solution.
In the case of solving \((x+10)^2=11\) using the square root property, we involve a few specific steps:

• First, understand that applying the square root is a mathematical operation to simplify the equation.
• The operation of taking the square root is necessary here to "get rid" of the square on the left side of the equation.
• Once the square root has been taken, other operations, like subtraction in this instance, are carried out to isolate the variable \(x\).

Mastery of these basic operations is essential for anyone tackling algebraic equations since each operation serves a unique purpose in unraveling different types of mathematical problems.