Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - 10-5 Graphing Square Root Functions - Mixed Review - Page 631: 71


There will be no solution for the given equation.

Work Step by Step

Given Infomation : $\sqrt 2r - 3 = r$ ------- eq1 By squaring the eq 1 we get $2r- 3= r^{2}$ ---- eq 2 By simplifying the eq 2 on both sides we get, $0=r^{2}-2r+3$ ----eq3 Now, by factoring the eq 3 $0=(r-3)(r+1)$ By using the zero product property $r-3=$ -> r=3 $r+1$ -> r=-1 Now, By checking At r=3, $\sqrt 2(3)-3=3$ $\sqrt 3\ne3$ r=-1 $\sqrt 2(-1)-3=-1$ $\sqrt -5\ne-1$ So Both are extraneous solutions. Hence there are 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.