gibsonj338
Sep 28, 2018

#2**0 **

\(x+\frac{5}{x}-1+x+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-1+\frac{8}{x}+1=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}-0+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{5}{x}+\frac{8}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1\)

\(2x+\frac{13}{x}-8x=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=8x+\frac{9}{{x}^{2}}-1-8x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0x\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1-0\)

\(-6x+\frac{13}{x}=\frac{9}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{9}{{x}^{2}}-1-\frac{9}{{x}^{2}}\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=\frac{0}{{x}^{2}}-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=0-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}=-1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=-1+1\)

\(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1=0\)

\({x}^{2}(-6x+\frac{13}{x}-\frac{9}{{x}^{2}}+1)=0\times{x}^{2}\)

\(-6{x}^{3}+\frac{13{x}^{2}}{x}-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-\frac{9{x}^{2}}{{x}^{2}}+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+1{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\times{x}^{2}\)

\(-6{x}^{3}+13x-9+{x}^{2}=0\)

\(-6{x}^{3}+{x}^{2}+13x-9=0\)

Since you cannot factor any further, the only way I know of to finish solving this equation is by graphing.

Click on the following link to view the graph: https://www.desmos.com/calculator/srwqsdcfr8

When looking at the graph, you find that x ≈ -1.669416

gibsonj338Sep 16, 2017