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Ch. R - Algebra Review
Lial - Trigonometry 12th Edition
Lial12th EditionTrigonometryISBN: 9780136552161Not the one you use?Change textbook
Chapter 1, Problem 8

CONCEPT PREVIEW Use choices A–D to answer each question.


A. 3x² - 17x - 6 = 0
B.(2x + 5)² = 7
C. x² + x = 12
D. (3x - 1) (x - 7) = 0


Which quadratic equation is set up for direct use of the square root property? Solve it.

Verified step by step guidance
1
Identify the quadratic equation that can be solved directly by applying the square root property. This property is used when the equation is in the form \( (ax + b)^2 = c \), where you can take the square root of both sides.
Look at each option and check if it matches the form \( (ax + b)^2 = c \). Option B is \( (2x + 5)^2 = 7 \), which fits this form perfectly.
Apply the square root property by taking the square root of both sides: \( 2x + 5 = \pm \sqrt{7} \).
Solve for \( x \) by isolating it: subtract 5 from both sides to get \( 2x = -5 \pm \sqrt{7} \), then divide both sides by 2 to find \( x = \frac{-5 \pm \sqrt{7}}{2} \).
Express the solution as two values corresponding to the \( \pm \) sign, representing the two possible solutions for \( x \).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Square Root Property

The square root property states that if x² = k, then x = ±√k. This method is used to solve quadratic equations that can be written in the form (expression)² = constant, allowing direct extraction of the square root without factoring or using the quadratic formula.
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Identifying Quadratic Equations Suitable for the Square Root Property

A quadratic equation suitable for the square root property is one where the variable term is isolated and squared, such as (ax + b)² = c. Recognizing this form helps avoid unnecessary steps like factoring or completing the square.
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Solving Quadratic Equations Using the Square Root Property

To solve using the square root property, isolate the squared term, take the square root of both sides, and include both positive and negative roots. Then solve the resulting linear equations to find all possible solutions.
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