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Ch. 1 - Equations and Inequalities
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
All textbooksBlitzer 8th EditionCh. 1 - Equations and InequalitiesProblem 88
Chapter 2, Problem 88

Solve each radical equation in Exercises 88–89. √ (2x-3) + x = 3

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Step 1: Isolate the radical expression on one side of the equation. Subtract x from both sides to get: √(2x - 3) = 3 - x.
Step 2: Eliminate the square root by squaring both sides of the equation. This gives: (√(2x - 3))² = (3 - x)². Simplify to get: 2x - 3 = (3 - x)(3 - x).
Step 3: Expand the right-hand side of the equation. Use the distributive property to expand (3 - x)(3 - x), which results in: 2x - 3 = 9 - 6x + x².
Step 4: Rearrange the equation into standard quadratic form. Combine all terms on one side of the equation to get: x² - 8x + 12 = 0.
Step 5: Solve the quadratic equation using factoring, the quadratic formula, or completing the square. After finding the solutions, substitute them back into the original equation to check for extraneous solutions introduced by squaring.

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Key Concepts

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

Radical Equations

Radical equations are equations that involve a variable within a radical (square root, cube root, etc.). To solve these equations, one typically isolates the radical on one side and then squares both sides to eliminate the radical. This process may introduce extraneous solutions, so it's essential to check all potential solutions in the original equation.
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Isolating the Variable

Isolating the variable is a fundamental algebraic technique used to solve equations. This involves rearranging the equation to get the variable on one side and all other terms on the opposite side. In the context of radical equations, isolating the radical before squaring both sides is crucial for correctly solving the equation.
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Extraneous Solutions

Extraneous solutions are solutions that emerge from the process of solving an equation but do not satisfy the original equation. This is particularly common in radical equations, where squaring both sides can introduce solutions that are not valid. Therefore, it is important to substitute any found solutions back into the original equation to verify their validity.
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Related Practice
Textbook Question

The equations in Exercises 79–90 combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 4/(x - 2) + 3/(x + 5) = 7/(x + 5)(x - 2)

Textbook Question

In Exercises 59–94, solve each absolute value inequality. 5 > |4 - x|

Textbook Question

Solve each equation in Exercises 83–108 by the method of your choice. x2−2x=1x^2 - 2x = 1

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Textbook Question

The equations in Exercises 79–90 combine the types of equations we have discussed in this section. Solve each equation. Then state whether the equation is an identity, a conditional equation, or an inconsistent equation. 4x/(x + 3) - 12/(x - 3) = (4x2 + 36)/(x2 - 9)

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Textbook Question

In Exercises 85–90, find the x-intercepts of the graph of each equation. Then use the x-intercepts to match the equation with its graph. [The graphs are labeled (a) through (f).]

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y=x−2−x−1−6y = x^{-2} - x^{-1} - 6

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Textbook Question

In Exercises 59–94, solve each absolute value inequality. 1 < |2 - 3x|

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