Solve each equation in Exercises 15–34 by the square root property. (2x + 8)^2 = 27

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 \(a = 2x + 8\) and \(b = 27\).
2. Take the square root of both sides of the equation: \(2x + 8 = \pm \sqrt{27}\).
3. Simplify \(\sqrt{27}\) to \(3\sqrt{3}\) because \(27 = 9 \times 3\) and \(\sqrt{9} = 3\).
4. Set up two separate equations to solve for \(x\): \(2x + 8 = 3\sqrt{3}\) and \(2x + 8 = -3\sqrt{3}\).
5. For each equation, isolate \(x\) by first subtracting 8 from both sides, then dividing by 2.

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 results in two possible equations: ax + b = √c and ax + b = -√c. This property is essential for solving equations that involve squares.

Isolating the Variable

Isolating the variable involves rearranging an equation to get the variable on one side and all other terms on the opposite side. In the context of the square root property, this means simplifying the equation to the form (ax + b)^2 = c before applying the square root. This step is crucial for correctly applying the square root property.

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 drawn from the algebraic manipulation.