Question Number 104588 by M±th+et+s last updated on 22/Jul/20 | ||
$${solve}\:{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{96}{x}−\mathrm{24}}{\mathrm{12}{x}+\mathrm{5}}=\sqrt{−\mathrm{144}{x}^{\mathrm{2}} +\mathrm{72}{x}+\mathrm{7}} \\ $$ | ||
Commented by M±th+et+s last updated on 22/Jul/20 | ||
$${thanks}\:{both}\:{for}\:{solutions} \\ $$ | ||
Answered by ajfour last updated on 22/Jul/20 | ||
$$\mathrm{8}−\left(\frac{\mathrm{64}}{\mathrm{12}{x}+\mathrm{5}}\right)=\sqrt{\mathrm{16}−\left(\mathrm{12}{x}−\mathrm{3}\right)^{\mathrm{2}} } \\ $$$${let}\:\:\mathrm{12}{x}−\mathrm{3}={t}\:\:\:{with}\:\:\:−\mathrm{4}\leqslant{t}\leqslant\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{8}−\frac{\mathrm{64}}{{t}+\mathrm{8}}=\sqrt{\mathrm{16}−{t}^{\mathrm{2}} } \\ $$$$\Rightarrow\:\:\mathrm{4}−\sqrt{\mathrm{16}−{t}^{\mathrm{2}} }=\frac{\mathrm{64}}{{t}+\mathrm{8}}−\mathrm{4}\:\:\:\:\:....\left({i}\right) \\ $$$$\Rightarrow\:\:\:\frac{{t}^{\mathrm{2}} }{\mathrm{4}+\sqrt{\mathrm{16}−{t}^{\mathrm{2}} }}\:=\:\frac{\mathrm{64}}{{t}+\mathrm{8}}−\mathrm{4} \\ $$$$\Rightarrow\:\:\mathrm{4}+\sqrt{\mathrm{16}−{t}^{\mathrm{2}} }\:=\:\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{{t}+\mathrm{8}}{\mathrm{8}−{t}}\right)\:\:\:...\left({ii}\right) \\ $$$${Adding}\:\:\left({i}\right)\:\&\:\left({ii}\right) \\ $$$$\:\:\:\:\:\mathrm{8}=\frac{{t}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{{t}+\mathrm{8}}{\mathrm{8}−{t}}\right)+\mathrm{4}\left(\frac{\mathrm{8}−{t}}{{t}+\mathrm{8}}\right) \\ $$$${Now}\:{let}\:\:\frac{\mathrm{8}−{t}}{{t}+\mathrm{8}}\:=\:{s}\:\:\Rightarrow\:\:{t}=\mathrm{8}\left(\frac{\mathrm{1}−{s}}{{s}+\mathrm{1}}\right) \\ $$$$\Rightarrow\:\:\:\mathrm{8}=\mathrm{16}\left(\frac{\mathrm{1}−{s}}{{s}+\mathrm{1}}\right)^{\mathrm{2}} \:\:\left(\frac{\mathrm{1}}{{s}}\right)+\mathrm{4}{s} \\ $$$$\Rightarrow\:\:\:\:\mathrm{2}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{1}−{s}\right)^{\mathrm{2}} +{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{2}−{s}\right)=\mathrm{4}\left(\mathrm{1}−{s}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\left({s}+\mathrm{1}\right)^{\mathrm{2}} \left[\mathrm{1}−\left({s}−\mathrm{1}\right)^{\mathrm{2}} \right]=\mathrm{4}\left(\mathrm{1}−{s}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{s}\right)^{\mathrm{2}} }−\frac{\mathrm{4}}{\left(\mathrm{1}+{s}\right)^{\mathrm{2}} }\:=\:\mathrm{1} \\ $$$$...... \\ $$ | ||