Ch. 1 - Equations and Inequalities

Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook

Lial 13th Edition

Ch. 1 - Equations and Inequalities

Problem 59

Chapter 2, Problem 59

Solve each equation. √3x=√(5x+1)-1

Verified step by step guidance

Start with the given equation: $\sqrt{3x} = \sqrt{5x + 1} - 1$.

Isolate one of the square root terms by adding 1 to both sides: $\sqrt{3x} + 1 = \sqrt{5x + 1}$.

Square both sides of the equation to eliminate the square roots: $\left(\sqrt{3x} + 1\right)^2 = \left(\sqrt{5x + 1}\right)^2$.

Expand the left side using the formula $(a + b)^2 = a^2 + 2ab + b^2$: $3x + 2\sqrt{3x} + 1 = 5x + 1$.

Simplify the equation and isolate the remaining square root term, then square both sides again to solve for $x$.

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Key Concepts

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

Square Root Equations

Square root equations involve variables under a radical sign. To solve them, isolate the square root expression and then square both sides to eliminate the radical. This process may introduce extraneous solutions, so checking all solutions in the original equation is essential.

Isolating Variables

Isolating the variable means manipulating the equation to get the variable alone on one side. This step is crucial before squaring both sides to avoid complicating the equation further. Proper isolation helps simplify the solving process and reduces errors.

Checking for Extraneous Solutions

Squaring both sides of an equation can introduce solutions that do not satisfy the original equation. These are called extraneous solutions. Always substitute the found solutions back into the original equation to verify which ones are valid.

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