Question

$$(1+x+ { x }^{ 2 } )/(1-x+ { x }^{ 2 } )=62(1+x)/63(1-x)$$

Answer

x=-0.80685043334961,-0.0080329895019531

Solution


Simplify  \(62\times \frac{1+x}{63}(1-x)\)  to  \(\frac{62(1+x)(1-x)}{63}\).
\[\frac{1+x+{x}^{2}}{1-x+{x}^{2}}=\frac{62(1+x)(1-x)}{63}\]
Multiply both sides by \(63\).
\[\frac{63(1+x+{x}^{2})}{1-x+{x}^{2}}=62(1+x)(1-x)\]
Expand.
\[\frac{63(1+x+{x}^{2})}{1-x+{x}^{2}}=62-62{x}^{2}\]
Multiply both sides by \(1-x+{x}^{2}\).
\[63(1+x+{x}^{2})=(62-62{x}^{2})(1-x+{x}^{2})\]
Expand.
\[63+63x+63{x}^{2}=62-62x+62{x}^{2}-62{x}^{2}+62{x}^{3}-62{x}^{4}\]
Simplify  \(62-62x+62{x}^{2}-62{x}^{2}+62{x}^{3}-62{x}^{4}\)  to  \(62-62x+62{x}^{3}-62{x}^{4}\).
\[63+63x+63{x}^{2}=62-62x+62{x}^{3}-62{x}^{4}\]
Move all terms to one side.
\[63+63x+63{x}^{2}-62+62x-62{x}^{3}+62{x}^{4}=0\]
Simplify  \(63+63x+63{x}^{2}-62+62x-62{x}^{3}+62{x}^{4}\)  to  \(1+125x+63{x}^{2}-62{x}^{3}+62{x}^{4}\).
\[1+125x+63{x}^{2}-62{x}^{3}+62{x}^{4}=0\]
No root was found algebraically. However, the following root(s) were found by numerical methods.
\[x=-0.806850,-0.008033\]