Solve each equation in Exercises 15–34 by the square root property. (3x + 2)^2 = 9

Verified step by step guidance

1. Start by applying the square root property, which states that if \((a)^2 = b\), then \(a = \pm \sqrt{b}\). In this case, set \((3x + 2)^2 = 9\) to \(3x + 2 = \pm \sqrt{9}\).
2. Calculate the square root of 9, which is 3. So, you have two equations: \(3x + 2 = 3\) and \(3x + 2 = -3\).
3. Solve the first equation \(3x + 2 = 3\) by subtracting 2 from both sides to isolate the term with \(x\).
4. Solve the second equation \(3x + 2 = -3\) by subtracting 2 from both sides to isolate the term with \(x\).
5. For both equations, divide by 3 to solve for \(x\). This will give you the two possible solutions for \(x\).

Key Concepts

Square Root Property

The square root property states that if a quadratic equation is in the form (ax + b)^2 = c, then the solutions can be found by taking the square root of both sides. This leads to two possible equations: ax + b = √c and ax + b = -√c. This property is essential for solving equations that can be expressed as perfect squares.

Isolating the Variable

Isolating the variable involves rearranging the equation to get the variable on one side and the constants on the other. In the context of the square root property, this often means first simplifying the equation to the form (ax + b)^2 = c before applying the square root. This step is crucial for accurately finding the values of the variable.

Extraneous Solutions

Extraneous solutions are solutions that emerge from the algebraic process but do not satisfy the original equation. When using the square root property, it is important to check each potential solution by substituting it back into the original equation to ensure it is valid. This helps avoid incorrect conclusions that may arise from the algebraic manipulation.