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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 19

Solve each radical equation in Exercises 11–30. Check all proposed solutions. 2x+19−8=x\(\sqrt{2x + 19}\) - 8 = x

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Start by isolating the square root term on one side of the equation. Add 8 to both sides to get: \(\sqrt{2x + 19} = x + 8\).
Next, square both sides of the equation to eliminate the square root. This gives: \((\sqrt{2x + 19})^2 = (x + 8)^2\), which simplifies to \(2x + 19 = (x + 8)^2\).
Expand the right side of the equation: \((x + 8)^2 = x^2 + 16x + 64\). So the equation becomes \(2x + 19 = x^2 + 16x + 64\).
Rearrange the equation to set it equal to zero by subtracting \(2x + 19\) from both sides: \(0 = x^2 + 16x + 64 - 2x - 19\), which simplifies to \(0 = x^2 + 14x + 45\).
Solve the quadratic equation \(x^2 + 14x + 45 = 0\) using factoring, completing the square, or the quadratic formula. After finding the solutions, check each one by substituting back into the original equation to verify they do not produce extraneous results.

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