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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
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Tutti i libri di testoBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblema 101
Capitolo 2, Problema 101

In Exercises 101–106, solve each equation. ∣x−5∣=2\(\left\)|\(\sqrt{x}\)-5\(\right\)|=2

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Start by understanding the equation: \(|\sqrt{x} - 5| = 2\). The absolute value means the expression inside can be either 2 or -2.
Set up two separate equations to remove the absolute value: \(\sqrt{x} - 5 = 2\) and \(\sqrt{x} - 5 = -2\).
Solve the first equation: add 5 to both sides to get \(\sqrt{x} = 7\), then square both sides to eliminate the square root, resulting in \(x = 7^2\).
Solve the second equation: add 5 to both sides to get \(\sqrt{x} = 3\), then square both sides to eliminate the square root, resulting in \(x = 3^2\).
Check both solutions in the original equation to ensure they do not produce extraneous results, since squaring can introduce invalid solutions.

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Absolute Value Equations

An absolute value equation involves expressions within absolute value bars, which represent the distance from zero on the number line. To solve |A| = B, where B ≥ 0, split it into two cases: A = B and A = -B. This approach helps find all possible solutions that satisfy the original equation.
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Categorizing Linear Equations

Square Root Function

The square root function, √x, returns the non-negative number whose square is x. It is defined only for x ≥ 0 in the real numbers. Understanding its domain is crucial when solving equations involving √x to ensure solutions are valid.
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Imaginary Roots with the Square Root Property

Solving Equations with Radicals and Absolute Values

When solving equations that combine radicals and absolute values, isolate the absolute value expression first, then consider the domain restrictions from the radical. After splitting the absolute value into two cases, solve each resulting equation carefully, checking for extraneous solutions.
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Solving Logarithmic Equations
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Exercises 100–102 will help you prepare for the material covered in the next section. Factor: x2−6x+9x^2 - 6x + 9

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In Exercises 91–100, find all values of x satisfying the given conditions.y1=6(2xx−3)2,y2=5(2xx−3),andy1 exceeds y2 by 6.y_1 = 6 \(\left\)( \(\frac{2x}{x - 3}\) \(\right\))^2, \(\quad\) y_2 = 5 \(\left\)( \(\frac{2x}{x - 3}\) \(\right\)), \(\quad\) \(\text{and}\) \(\quad\) y_1 \(\text{ exceeds }\) y_2 \(\text{ by }\) 6.

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