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Question Number 25477 by Mcfaithal last updated on 11/Dec/17

((1+x)/2)=x−3+(5/x) LCM=( 2)(x) multiply each step by 2x (2)(x)(((1+x)/2))=(2)(x)(x−3)+(2)(x)((5/x)) (x)(1+x)=2x(x−3)+(2)(5) x+x^2 =2x^2 −6x+10 move all equations to LHS 2x^2 −x^2 −6x−x+10=0 x^2 −5x+10=0 x=((5±(√(25−40)))/2)=((5±(√(−15)))/2)=(1/2)(5±i(√(15)))

$$\frac{\mathrm{1}+{x}}{\mathrm{2}}={x}−\mathrm{3}+\frac{\mathrm{5}}{{x}} \\ $$$${LCM}=\left(\:\mathrm{2}\right)\left({x}\right) \\ $$$${multiply}\:{each}\:{step}\:{by}\:\mathrm{2}{x} \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\left({x}\right)\left(\frac{\mathrm{1}+{x}}{\mathrm{2}}\right)=\left(\mathrm{2}\right)\left({x}\right)\left({x}−\mathrm{3}\right)+\left(\mathrm{2}\right)\left({x}\right)\left(\frac{\mathrm{5}}{{x}}\right) \\ $$$$ \\ $$$$\left({x}\right)\left(\mathrm{1}+{x}\right)=\mathrm{2}{x}\left({x}−\mathrm{3}\right)+\left(\mathrm{2}\right)\left(\mathrm{5}\right)\:\:\:\:\:\:\:\:\:\:\: \\ $$$${x}+{x}^{\mathrm{2}} =\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{10} \\ $$$${move}\:{all}\:{equations}\:{to}\:{LHS} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −{x}^{\mathrm{2}} −\mathrm{6}{x}−{x}+\mathrm{10}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{10}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{40}}}{\mathrm{2}}=\frac{\mathrm{5}\pm\sqrt{−\mathrm{15}}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{5}\pm{i}\sqrt{\mathrm{15}}\right) \\ $$

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