Question

25x^-4-99x^-2-4=0 How do I solve for this?

Answer 1

Given

$25x^{-4} - 99x^{-2} - 4 = 0$

consider substituting $y=x^{-2}$ to get a proper quadratic equation,

$25y^2 - 99y - 4 = 0$

Solve for $y$ ; we can factorize to get

$(25y + 1) (y - 4) = 0$

$25y+1 = 0 \text{ or } y - 4 = 0$

$y = -\dfrac1{25} \text{ or }y = 4$

Solve for $x$ :

$x^{-2} = -\dfrac1{25} \text{ or }x^{-2} = 4$

The first equation has no real solution, since $x^{-2} = \frac1{x^2} > 0$ for all non-zero $x$. Proceeding with the second equation, we get

$x^{-2} = 4 \implies x^2 = \dfrac14 \implies x = \pm\sqrt{\dfrac14} = \boxed{\pm \dfrac12}$

If we want to find all complex solutions, we take $i=\sqrt{-1}$ so that the first equation above would have led us to

$x^{-2} = -\dfrac1{25} \implies x^2 = -25 \implies x = \pm\sqrt{-25} = \pm5i$