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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 2, Problem 23

Solve each equation in Exercises 15–34 by the square root property. 3(x4)2=153(x - 4)^2 = 15

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1
Start with the given equation: \(3(x - 4)^2 = 15\).
Divide both sides of the equation by 3 to isolate the squared term: \((x - 4)^2 = \frac{15}{3}\).
Simplify the right side: \((x - 4)^2 = 5\).
Apply the square root property, which states that if \(a^2 = b\), then \(a = \pm \sqrt{b}\). So, \(x - 4 = \pm \sqrt{5}\).
Solve for \(x\) by adding 4 to both sides: \(x = 4 \pm \sqrt{5}\).

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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 property is used to solve equations where a variable is squared and isolated, allowing you to take the square root of both sides to find the variable's values.
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Isolating the Squared Term

Before applying the square root property, the equation must be manipulated so that the squared term stands alone on one side. This often involves dividing or simplifying the equation to isolate the expression with the exponent.
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Simplifying Radicals

After taking the square root, the result may include a radical expression. Simplifying radicals involves reducing the square root to its simplest form, which helps in expressing the solution clearly and accurately.
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