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


Sagot :

Given

[tex]25x^{-4} - 99x^{-2} - 4 = 0[/tex]

consider substituting [tex]y=x^{-2}[/tex] to get a proper quadratic equation,

[tex]25y^2 - 99y - 4 = 0[/tex]

Solve for [tex]y[/tex] ; we can factorize to get

[tex](25y + 1) (y - 4) = 0[/tex]

[tex]25y+1 = 0 \text{ or } y - 4 = 0[/tex]

[tex]y = -\dfrac1{25} \text{ or }y = 4[/tex]

Solve for [tex]x[/tex] :

[tex]x^{-2} = -\dfrac1{25} \text{ or }x^{-2} = 4[/tex]

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

[tex]x^{-2} = 4 \implies x^2 = \dfrac14 \implies x = \pm\sqrt{\dfrac14} = \boxed{\pm \dfrac12}[/tex]

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

[tex]x^{-2} = -\dfrac1{25} \implies x^2 = -25 \implies x = \pm\sqrt{-25} = \pm5i[/tex]