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q-1-q-2-x-0-8-x-2-x-




Question Number 27334 by darkalestero1@gmail.com last updated on 05/Jan/18
(q_1 /q_2 )=((x/(0.8−x)))^2     ; x=?
$$\frac{\mathrm{q}_{\mathrm{1}} }{\mathrm{q}_{\mathrm{2}} }=\left(\frac{\mathrm{x}}{\mathrm{0}.\mathrm{8}−\mathrm{x}}\right)^{\mathrm{2}} \:\:\:\:;\:\boldsymbol{\mathrm{x}}=? \\ $$
Answered by mrW1 last updated on 05/Jan/18
(x/(0.8−x))=±(√(q_1 /q_2 ))  ((0.8−x)/x)=±(√(q_2 /q_1 ))  ((0.8)/x)=1±(√(q_2 /q_1 ))  ⇒x=((0.8)/(1±(√(q_2 /q_1 ))))
$$\frac{{x}}{\mathrm{0}.\mathrm{8}−{x}}=\pm\sqrt{\frac{{q}_{\mathrm{1}} }{{q}_{\mathrm{2}} }} \\ $$$$\frac{\mathrm{0}.\mathrm{8}−{x}}{{x}}=\pm\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }} \\ $$$$\frac{\mathrm{0}.\mathrm{8}}{{x}}=\mathrm{1}\pm\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }} \\ $$$$\Rightarrow{x}=\frac{\mathrm{0}.\mathrm{8}}{\mathrm{1}\pm\sqrt{\frac{{q}_{\mathrm{2}} }{{q}_{\mathrm{1}} }}} \\ $$

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