Help finding the real solutions in this equation

[ProdigaI]

x^-2 - 8x^-1 + 12 = 0

x to the negative 2 power minus 8x to the negative 1 power plus 12 equals 0

Thanks

 

[soroban]

Hello, ProdigaI!

\(\displaystyle \text{Solve: }\;x^{-2} - 8x^{-1} + 12 \:=\: 0\)

\(\displaystyle \text{Multiply by }x^2\!:\quad 1 - 8x + 12x^2 \:=\:0\quad\Rightarrow\quad 12x^2 - 8x + 1 \:=\:0\)

. . \(\displaystyle \text{which factors: }\;(2x - 1)(6x - 1) \:=\;0\)

. . \(\displaystyle \text{and has roots: }\;x \:=\:\tfrac{1}{2},\:\tfrac{1}{6}\)

 

[Deleted member 4993]

x^-2 - 8x^-1 + 12 = 0

x to the negative 2 power minus 8x to the negative 1 power plus 12 equals 0

Thanks

substitute:

u = 1/x

then

u^2 - 8u + 12 = 0

(u -6)(u-2) = 0

u = 2 converts to x = 1/2

and

u = 6 converts to x = 1/6