Question

$$ln=\frac{x^{2}+x+1}{x^{2}-x+1}$$

Answer

x=IM/sqrt(2),-IM/sqrt(2)

Solution


Simplify  \(1\times {x}^{2}\)  to  \({x}^{2}\).
\[ln={x}^{2}+x+{x}^{2}-x+1\]
Simplify  \({x}^{2}+x+{x}^{2}-x+1\)  to  \(2{x}^{2}+1\).
\[ln=2{x}^{2}+1\]
Subtract \(1\) from both sides.
\[ln-1=2{x}^{2}\]
Divide both sides by \(2\).
\[\frac{ln-1}{2}={x}^{2}\]
Take the square root of both sides.
\[\pm \sqrt{\frac{ln-1}{2}}=x\]
Simplify  \(\sqrt{\frac{ln-1}{2}}\)  to  \(\frac{\sqrt{ln-1}}{\sqrt{2}}\).
\[\pm \frac{\sqrt{ln-1}}{\sqrt{2}}=x\]
Rewrite \(ln-1\) in the form \({a}^{2}-{b}^{2}\), where \(a=0\) and \(b=1\).
\[\pm \frac{\sqrt{{0}^{2}-{1}^{2}}}{\sqrt{2}}=x\]
Simplify  \({0}^{2}\)  to  \(0\).
\[\pm \frac{\sqrt{0-{1}^{2}}}{\sqrt{2}}=x\]
Simplify  \({1}^{2}\)  to  \(1\).
\[\pm \frac{\sqrt{0-1}}{\sqrt{2}}=x\]
Simplify  \(0-1\)  to  \(-1\).
\[\pm \frac{\sqrt{-1}}{\sqrt{2}}=x\]
Simplify  \(\sqrt{-1}\)  to  \(\sqrt{1}\imath \).
\[\pm \frac{\sqrt{1}\imath }{\sqrt{2}}=x\]
Simplify  \(\sqrt{1}\)  to  \(1\).
\[\pm \frac{1\times \imath }{\sqrt{2}}=x\]
Simplify  \(1\times \imath \)  to  \(\imath \).
\[\pm \frac{\imath }{\sqrt{2}}=x\]
Switch sides.
\[x=\pm \frac{\imath }{\sqrt{2}}\]