Given
![25x^(-4) - 99x^(-2) - 4 = 0](https://img.qammunity.org/2023/formulas/mathematics/high-school/dmho9h2wed3m3l2qcwea755sp9ujh1xq4o.png)
consider substituting
to get a proper quadratic equation,
![25y^2 - 99y - 4 = 0](https://img.qammunity.org/2023/formulas/mathematics/high-school/eq9ap6qpmiqcafclf0wc3an0zeu6dmp70z.png)
Solve for
; we can factorize to get
![(25y + 1) (y - 4) = 0](https://img.qammunity.org/2023/formulas/mathematics/high-school/wlh6oz3osz22umde1zb1wvni63hrdowem4.png)
![25y+1 = 0 \text{ or } y - 4 = 0](https://img.qammunity.org/2023/formulas/mathematics/high-school/hlw8wdgi3r4f3agppyf0s0k9bmo7waojnm.png)
![y = -\frac1{25} \text{ or }y = 4](https://img.qammunity.org/2023/formulas/mathematics/high-school/gfx0k9rrbwvlfs1gve4ef64hhnycqhrn4p.png)
Solve for
:
![x^(-2) = -\frac1{25} \text{ or }x^(-2) = 4](https://img.qammunity.org/2023/formulas/mathematics/high-school/k1y03yyphd4mp56pjetu3d5l2e28shulbw.png)
The first equation has no real solution, since
for all non-zero
. Proceeding with the second equation, we get
![x^(-2) = 4 \implies x^2 = \frac14 \implies x = \pm√(\frac14) = \boxed{\pm \frac12}](https://img.qammunity.org/2023/formulas/mathematics/high-school/g7gab6xp8xvipdxzyoh1rjy4ilj61hacfb.png)
If we want to find all complex solutions, we take
so that the first equation above would have led us to
![x^(-2) = -\frac1{25} \implies x^2 = -25 \implies x = \pm√(-25) = \pm5i](https://img.qammunity.org/2023/formulas/mathematics/high-school/gxemivviu81f041rwok8klni5cds6pdbgq.png)