How do you solve #81x ^ { - 4} - 323x ^ { - 2} - 4= 0#?

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Answer 1 · 2017-03-17T06:59:51

#x=+-1/2# or #x=+-9i#

To solve the equation #81x^(-4)-323x^(-2)-4=0# or

#81/x^4-323/x^2-4=0#, let us assume #y=1/x^2#

Hence #81y^2-323y-4=0#

or #81y^2-324y+y-4=0#

or #81y(y-4)+1(y-4)=0#

or #(81y+1)(y-4)=0#

Hence either #y=-1/81# or #y=4#

if #y=-1/81#, #1/x^2=-1/81# #x^2=-81# and #x=+-9i#

and if #y=4#, #1/x^2=4# #x^2=1/4# and #x=+-1/2#