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Question Number 35338 by jasno91 last updated on 18/May/18

((14x^2 +16)/(21))−((2x^2 +8)/(8x^2 −11))=((2x^2 )/3)

$$\frac{\mathrm{14}{x}^{\mathrm{2}} +\mathrm{16}}{\mathrm{21}}−\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{8}}{\mathrm{8}{x}^{\mathrm{2}} −\mathrm{11}}=\frac{\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 18/May/18

Let x^2 =y ((14y+16)/(21))−((2y+8)/(8y−11))=((2y)/3) (14y+16)(8y−11)−21(2y+8)=14y(8y−11) 112y^2 ^(×) −154y^(×) +128y−176=112y^2 ^(×) −154y^(×) 128y−176=0 y=((176)/(128))=((11)/8) x^2 =((11)/8) x=±(√((11)/8))=((±(√(11)))/(2(√2)))

$$\mathrm{Let}\:{x}^{\mathrm{2}} ={y} \\ $$$$\frac{\mathrm{14}{y}+\mathrm{16}}{\mathrm{21}}−\frac{\mathrm{2}{y}+\mathrm{8}}{\mathrm{8}{y}−\mathrm{11}}=\frac{\mathrm{2}{y}}{\mathrm{3}} \\ $$$$\left(\mathrm{14}{y}+\mathrm{16}\right)\left(\mathrm{8}{y}−\mathrm{11}\right)−\mathrm{21}\left(\mathrm{2}{y}+\mathrm{8}\right)=\mathrm{14}{y}\left(\mathrm{8}{y}−\mathrm{11}\right) \\ $$$$\overset{×} {\mathrm{112}{y}^{\mathrm{2}} }\overset{×} {−\mathrm{154}{y}}+\mathrm{128}{y}−\mathrm{176}=\overset{×} {\mathrm{112}{y}^{\mathrm{2}} }\overset{×} {−\mathrm{154}{y}} \\ $$$$\mathrm{128}{y}−\mathrm{176}=\mathrm{0} \\ $$$${y}=\frac{\mathrm{176}}{\mathrm{128}}=\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{11}}{\mathrm{8}} \\ $$$${x}=\pm\sqrt{\frac{\mathrm{11}}{\mathrm{8}}}=\frac{\pm\sqrt{\mathrm{11}}}{\mathrm{2}\sqrt{\mathrm{2}}} \\ $$

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