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Question Number 55240 by Otchere Abdullai last updated on 19/Feb/19

$$makerthesubjectoftherelation$$$$m=\frac{4\sqrt{u+r}}{v-r}$$

Answered by MJS last updated on 20/Feb/19

$$m(v-r)=4\sqrt{u+r}$$$$\mathrm{squaring}\mathrm{both}\mathrm{sides}$$$$m^2{(v-r)}^2=16(u+r)$$$$\mathrm{expand}\mathrm{for}r$$$$m^2{r}^2-2(m^{2}v+8)r+m^2{v}^2-16u=0$$$$\mathrm{solve}\mathrm{with}\mathrm{formula}\mathrm{for}{2}^{\mathrm{nd}}\mathrm{degree}$$$$r=\frac{8}{m^2}+v\pm \frac{4\sqrt{m^2(u+v)+4}}{m^2}$$

Commented by Otchere Abdullai last updated on 20/Feb/19

$$Thankyoumjssir!$$

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