Solve for f
f=\frac{1}{2}-\frac{1}{2x^{2}}
x\neq 0
Solve for x (complex solution)
x=-i\left(2f-1\right)^{-\frac{1}{2}}
x=i\left(2f-1\right)^{-\frac{1}{2}}\text{, }f\neq \frac{1}{2}
Solve for x
x=\sqrt{-\frac{1}{2f-1}}
x=-\sqrt{-\frac{1}{2f-1}}\text{, }f<\frac{1}{2}
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fx\times 2x=x^{2}-1
Multiply both sides of the equation by 2x.
fx^{2}\times 2=x^{2}-1
Multiply x and x to get x^{2}.
2x^{2}f=x^{2}-1
The equation is in standard form.
\frac{2x^{2}f}{2x^{2}}=\frac{x^{2}-1}{2x^{2}}
Divide both sides by 2x^{2}.
f=\frac{x^{2}-1}{2x^{2}}
Dividing by 2x^{2} undoes the multiplication by 2x^{2}.
f=\frac{1}{2}-\frac{1}{2x^{2}}
Divide x^{2}-1 by 2x^{2}.
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