Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 731: 27


The equation has no solutions.

Work Step by Step

$(1+\sqrt{x})^{2}+5(1+\sqrt{x})+6=0\qquad$...substitute $1+\sqrt{x}$ for $u$ $ u^{2}+5u+6=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-5\pm\sqrt{25-24}}{2}$ $u=\displaystyle \frac{-5\pm\sqrt{1}}{2}$ $u=\displaystyle \frac{-5\pm 1}{2}$ $u=\displaystyle \frac{-5+1}{2}=\frac{-4}{2}=-2$ or $u=\displaystyle \frac{-5-1}{2}=\frac{-6}{2}=-3$ Bring back $1+\sqrt{x}=u$. $1+\sqrt{x}=-2$ or $1+\sqrt{x}=-3$ $\sqrt{x}=-3$ or $\sqrt{x}=-2$ $\sqrt{x}$ has to be a positive number which is why we discard both solutions. The equation has no solutions.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.