Ch. 1 - Equations and Inequalities

Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook

Blitzer 8th Edition

Ch. 1 - Equations and Inequalities

Problem 13

Chapter 2, Problem 13

Solve each radical equation in Exercises 11–30. Check all proposed solutions. √(x + 3) = x - 3

Verified step by step guidance

Start by isolating the radical expression, which is already isolated as $ \sqrt{\!x + 3} = x - 3 $.

Square both sides of the equation to eliminate the square root. This gives $ (\sqrt{\!x + 3})^2 = (x - 3)^2 $, which simplifies to $ x + 3 = (x - 3)^2 $.

Expand the right side using the formula $ (a - b)^2 = a^2 - 2ab + b^2 $. So, $ (x - 3)^2 = x^2 - 6x + 9 $. Substitute this back to get $ x + 3 = x^2 - 6x + 9 $.

Rearrange the equation to set it equal to zero: $ 0 = x^2 - 6x + 9 - x - 3 $, which simplifies to $ 0 = x^2 - 7x + 6 $.

Solve the quadratic equation $ x^2 - 7x + 6 = 0 $ by factoring, completing the square, or using the quadratic formula. After finding the solutions, check each one in the original equation to ensure they do not produce extraneous solutions.

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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 involve variables inside a root, often a square root. To solve them, isolate the radical expression and then eliminate the root by raising both sides to the appropriate power, typically squaring for square roots.

Extraneous Solutions

When solving radical equations by squaring both sides, new solutions may appear that do not satisfy the original equation. These are called extraneous solutions and must be checked by substituting back into the original equation.

Domain Restrictions

The expression inside a square root must be non-negative for real solutions. This restriction limits the domain of the variable and must be considered before solving to avoid invalid solutions.

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