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make-r-the-subject-of-the-relation-m-4-u-r-v-r-




Question Number 55240 by Otchere Abdullai last updated on 19/Feb/19
make r the subject of the relation  m=((4(√(u+r)))/(v−r))
$${make}\:{r}\:{the}\:{subject}\:{of}\:{the}\:{relation} \\ $$$${m}=\frac{\mathrm{4}\sqrt{{u}+{r}}}{{v}−{r}} \\ $$
Answered by MJS last updated on 20/Feb/19
m(v−r)=4(√(u+r))  squaring both sides  m^2 (v−r)^2 =16(u+r)  expand for r  m^2 r^2 −2(m^2 v+8)r+m^2 v^2 −16u=0  solve with formula for 2^(nd)  degree  r=(8/m^2 )+v±((4(√(m^2 (u+v)+4)))/m^2 )
$${m}\left({v}−{r}\right)=\mathrm{4}\sqrt{{u}+{r}} \\ $$$$\mathrm{squaring}\:\mathrm{both}\:\mathrm{sides} \\ $$$${m}^{\mathrm{2}} \left({v}−{r}\right)^{\mathrm{2}} =\mathrm{16}\left({u}+{r}\right) \\ $$$$\mathrm{expand}\:\mathrm{for}\:{r} \\ $$$${m}^{\mathrm{2}} {r}^{\mathrm{2}} −\mathrm{2}\left({m}^{\mathrm{2}} {v}+\mathrm{8}\right){r}+{m}^{\mathrm{2}} {v}^{\mathrm{2}} −\mathrm{16}{u}=\mathrm{0} \\ $$$$\mathrm{solve}\:\mathrm{with}\:\mathrm{formula}\:\mathrm{for}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{degree} \\ $$$${r}=\frac{\mathrm{8}}{{m}^{\mathrm{2}} }+{v}\pm\frac{\mathrm{4}\sqrt{{m}^{\mathrm{2}} \left({u}+{v}\right)+\mathrm{4}}}{{m}^{\mathrm{2}} } \\ $$
Commented by Otchere Abdullai last updated on 20/Feb/19
Thank you mjs sir!
$${Thank}\:{you}\:{mjs}\:{sir}! \\ $$

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