A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
v2=√(2v+39)
anonymous
 one year ago
v2=√(2v+39)

This Question is Closed

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0You need to square both sides to solve for v. Then when you find solutions, check the solutions in the original equation because squaring both sides may introduce extraneous solutions.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[v2=\sqrt{2v+39}\] and how we know that all square roots are nonnegative, then \[v2\ge0\] \[v \ge2\] and the argument of a square root must be nonnegative, then \[2v+39\ge0\] \[v \le \frac{ 39 }{ 2 }\] now, in the first equation be square both sides \[(v2)^{2}=(\sqrt{2v+39})^{2}\] \[v ^{2} 4v+4=2v+39\] \[v ^{2} 2v35=0\] factoring \[(v+5)(v7)=0\] so v=5 or v=7 could be solutions, but at the start we said that v must be greater than 2, so the only solution to this equation is v=7
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.