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anonymous

  • one year ago

v-2=√(-2v+39)

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  1. mathstudent55
    • one year ago
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    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.

  2. anonymous
    • one year ago
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    \[v-2=\sqrt{-2v+39}\] 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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