anonymous
  • anonymous
v-2=√(-2v+39)
Algebra
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anonymous
  • anonymous
v-2=√(-2v+39)
Algebra
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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mathstudent55
  • mathstudent55
You 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
  • anonymous
\[v-2=\sqrt{-2v+39}\] and how we know that all square roots are non-negative, then \[v-2\ge0\] \[v \ge2\] and the argument of a square root must be non-negative, then \[-2v+39\ge0\] \[v \le \frac{ 39 }{ 2 }\] now, in the first equation be square both sides \[(v-2)^{2}=(\sqrt{-2v+39})^{2}\] \[v ^{2} -4v+4=-2v+39\] \[v ^{2} -2v-35=0\] factoring \[(v+5)(v-7)=0\] 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

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