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