Question Number 103685 by bemath last updated on 16/Jul/20
Answered by bobhans last updated on 16/Jul/20
$${x}−\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}\:=\:\sqrt{\mathrm{4}−{x}}\left\{\sqrt{\mathrm{3}−{x}}+\sqrt{\mathrm{2}−{x}}\right\} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}−\mathrm{2}{x}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}\:= \\ $$$$\left(\mathrm{4}−{x}\right)\left(\mathrm{5}−\mathrm{2}{x}\right)+\mathrm{2}\sqrt{\left(\mathrm{3}−{x}\right)\left(\mathrm{2}−{x}\right)} \\ $$$$\mathrm{4}{x}−\mathrm{7}\:= \\ $$$${x}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)}+\left(\mathrm{4}−{x}\right)\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)} \\ $$$$\mathrm{4}{x}−\mathrm{7}\:=\:\mathrm{4}\sqrt{\left(\mathrm{2}−{x}\right)\left(\mathrm{3}−{x}\right)} \\ $$$$\left(\mathrm{4}{x}−\mathrm{7}\right)^{\mathrm{2}} \:=\:\mathrm{16}\left(\mathrm{6}−\mathrm{5}{x}+{x}^{\mathrm{2}} \right) \\ $$$$−\mathrm{56}{x}+\mathrm{49}\:=\:\mathrm{96}−\mathrm{80}{x}\: \\ $$$$\mathrm{24}{x}\:=\:\mathrm{47}\:\Rightarrow\:{x}\:=\:\frac{\mathrm{47}}{\mathrm{24}}\:\bigstar \\ $$