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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

2x−5=(8/x)+10 2x^2 −5x=8+10x 2x^2 −15x−8=0 2x^2 −16x+x−8=0 2x(x−8)+1(x−8)=0 (x−8)(2x+1)=0 x=8,((−1)/2) g(8)=2×8−5=11 h(8)=(8/8)+10=11 f(8)=2^(8−5) +3=11 f(8)=g(8)=h(8) g(((−1)/2))=2×((−1)/2)−5=−6 h(((−1)/2))=(8/(−1))×2+10=−6 f(((−1)/2))=2^(((−1)/2)−5) +3 f(((−1)/2))≠g(((−1)/2)) f(((−1)/2))≠h(((−1)/2)) so x=8 f(8)=g(8)=h(8)

$$\mathrm{2}{x}−\mathrm{5}=\frac{\mathrm{8}}{{x}}+\mathrm{10} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}=\mathrm{8}+\mathrm{10}{x} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{15}{x}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{16}{x}+{x}−\mathrm{8}=\mathrm{0} \\ $$$$\mathrm{2}{x}\left({x}−\mathrm{8}\right)+\mathrm{1}\left({x}−\mathrm{8}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{8}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$${x}=\mathrm{8},\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$${g}\left(\mathrm{8}\right)=\mathrm{2}×\mathrm{8}−\mathrm{5}=\mathrm{11}\:\:\:{h}\left(\mathrm{8}\right)=\frac{\mathrm{8}}{\mathrm{8}}+\mathrm{10}=\mathrm{11} \\ $$$${f}\left(\mathrm{8}\right)=\mathrm{2}^{\mathrm{8}−\mathrm{5}} +\mathrm{3}=\mathrm{11} \\ $$$${f}\left(\mathrm{8}\right)={g}\left(\mathrm{8}\right)={h}\left(\mathrm{8}\right) \\ $$$${g}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}×\frac{−\mathrm{1}}{\mathrm{2}}−\mathrm{5}=−\mathrm{6} \\ $$$${h}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{8}}{−\mathrm{1}}×\mathrm{2}+\mathrm{10}=−\mathrm{6} \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)=\mathrm{2}^{\frac{−\mathrm{1}}{\mathrm{2}}−\mathrm{5}} +\mathrm{3} \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\neq{g}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\ $$$${f}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\neq{h}\left(\frac{−\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\boldsymbol{{so}}\:\boldsymbol{{x}}=\mathrm{8}\:\:\boldsymbol{{f}}\left(\mathrm{8}\right)=\boldsymbol{{g}}\left(\mathrm{8}\right)=\boldsymbol{{h}}\left(\mathrm{8}\right) \\ $$

Commented by I want to learn more last updated on 13/Nov/20

Thanks sir.

$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$

Commented by TANMAY PANACEA last updated on 14/Nov/20

most welcome

$${most}\:{welcome} \\ $$

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