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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 2, Problema 23

Solve each equation in Exercises 15–34 by the square root property. 3(x−4)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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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

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