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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

$$2x-5=\frac{8}{x}+10$$$$2x^2-5x=8+10x$$$$2x^2-15x-8=0$$$$2x^2-16x+x-8=0$$$$2x(x-8)+1(x-8)=0$$$$(x-8)(2x+1)=0$$$$x=8,\frac{-1}{2}$$$$g\left(8\right)=2\times 8-5=11h\left(8\right)=\frac{8}{8}+10=11$$$$f\left(8\right)=2^{8-5}+3=11$$$$f\left(8\right)=g\left(8\right)=h\left(8\right)$$$$g\left(\frac{-1}{2}\right)=2\times \frac{-1}{2}-5=-6$$$$h\left(\frac{-1}{2}\right)=\frac{8}{-1}\times 2+10=-6$$$$f\left(\frac{-1}{2}\right)=2^{\frac{-1}{2}-5}+3$$$$f\left(\frac{-1}{2}\right)\ne g\left(\frac{-1}{2}\right)$$$$f\left(\frac{-1}{2}\right)\ne h\left(\frac{-1}{2}\right)$$$$\mathit{s}\mathit{o}\mathit{x}=8\mathit{f}\left(8\right)=\mathit{g}\left(8\right)=\mathit{h}\left(8\right)$$

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

$$mostwelcome$$

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