Question Number 122010 by I want to learn more last updated on 13/Nov/20
Answered by TANMAY PANACEA last updated on 13/Nov/20
$$\mathrm{2}{x}−\mathrm{5}=\frac{\mathrm{8}}{{x}}+\mathrm{10} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{8}+\mathrm{10}{x} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{15}{x}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{16}{x}+{x}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({x}−\mathrm{8}\right)+\mathrm{1}\left({x}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{8}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{8},\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left(\mathrm{8}\right)=\mathrm{2}×\mathrm{8}−\mathrm{5}=\mathrm{11}\:\:\:{h}\left(\mathrm{8}\right)=\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{10}=\mathrm{11} \\ $$$${f}\left(\mathrm{8}\right)=\mathrm{2}^{\mathrm{8}−\mathrm{5}} +\mathrm{3}=\mathrm{11} \\ $$$${f}\left(\mathrm{8}\right)={g}\left(\mathrm{8}\right)={h}\left(\mathrm{8}\right) \\ $$$${g}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{2}}−\mathrm{5}=−\mathrm{6} \\ $$$${h}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{8}}{−\mathrm{1}}×\mathrm{2}+\mathrm{10}=−\mathrm{6} \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}^{\frac{−\mathrm{1}}{\mathrm{2}}−\mathrm{5}} +\mathrm{3} \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\neq{g}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\neq{h}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{x}}=\mathrm{8}\:\:\boldsymbol{{f}}\left(\mathrm{8}\right)=\boldsymbol{{g}}\left(\mathrm{8}\right)=\boldsymbol{{h}}\left(\mathrm{8}\right) \\ $$
Commented by I want to learn more last updated on 13/Nov/20
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Commented by TANMAY PANACEA last updated on 14/Nov/20
$${most}\:{welcome} \\ $$