Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.4 - Equations Quadratic in Form - Exercise Set - Page 637: 62

Answer

Does not make sense. Checking is essential if at any point in the solution process both sides of an equation are raised to an even power. This is because extraneous solutions that are not solutions of the given equation may have been introduced. For instance: Solve: $2x - \sqrt{x} - 10 = 0$. $2x - x^{\frac{1}{2}}- 10 = 0$ Let $u =x^{\frac{1}{2}}$ $2(x^{\frac{1}{2}})^2 - x^{\frac{1}{2}}- 10 = 0$ $2u^2 - u - 10 = 0$ $(2u - 5)(u + 2) = 0$ $2u - 5 = 0$ or $u + 2 = 0$ $u =\frac{5}{2}$ or $u = -2$ Use the original substitution, $u =x^{\frac{1}{2}}$, to solve for $x$. $x^{\frac{1}{2}}=\frac{5}{2}$ or $x^{\frac{1}{2}}=-2$ Solve by squaring both sides. $(x^{\frac{1}{2}})^2=(\frac{5}{2})^2$ or $(x^{\frac{1}{2}})^2=(-2)^2$ $x =\frac{25}{4}$ or $x = 4$ Since both sides of the equations are squared, checking is required. Check $\frac{25}{4}$: $2x - \sqrt{x} - 10 = 0$ $2(\frac{25}{4}) - \sqrt{\frac{25}{4}} - 10 = 0$ $\frac{25}{2} -\frac{5}{2} - 10 = 0$ $\frac{20}{2}-10 = 0$ $0=0$ TRUE Check $4$: $2x - \sqrt{x} - 10 = 0$ $2(4) - \sqrt{4} - 10 = 0$ $8-2-10=0$ $6-10=0$ $-4=0$ FALSE The check indicates that $4$ is not a solution but an extraneous solution.

Work Step by Step

Checking is essential if at any point in the solution process both sides of an equation are raised to an even power. This is because extraneous solutions that are not solutions of the given equation may have been introduced. For instance: Solve: $2x - \sqrt{x} - 10 = 0$. $2x - x^{\frac{1}{2}}- 10 = 0$ Let $u =x^{\frac{1}{2}}$ $2(x^{\frac{1}{2}})^2 - x^{\frac{1}{2}}- 10 = 0$ $2u^2 - u - 10 = 0$ $(2u - 5)(u + 2) = 0$ $2u - 5 = 0$ or $u + 2 = 0$ $u =\frac{5}{2}$ or $u = -2$ Use the original substitution, $u =x^{\frac{1}{2}}$, to solve for $x$. $x^{\frac{1}{2}}=\frac{5}{2}$ or $x^{\frac{1}{2}}=-2$ Solve by squaring both sides. $(x^{\frac{1}{2}})^2=(\frac{5}{2})^2$ or $(x^{\frac{1}{2}})^2=(-2)^2$ $x =\frac{25}{4}$ or $x = 4$ Since both sides of the equations are squared, checking is required. Check $\frac{25}{4}$: $2x - \sqrt{x} - 10 = 0$ $2(\frac{25}{4}) - \sqrt{\frac{25}{4}} - 10 = 0$ $\frac{25}{2} -\frac{5}{2} - 10 = 0$ $\frac{20}{2}-10 = 0$ $0=0$ TRUE Check $4$: $2x - \sqrt{x} - 10 = 0$ $2(4) - \sqrt{4} - 10 = 0$ $8-2-10=0$ $6-10=0$ $-4=0$ FALSE The check indicates that $4$ is not a solution but an extraneous solution.
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