Question

Solve: 4x^-4 + 1 = 5x^-2

Answer 1

\[4/x^4+1=5/x^2\]\[4+x^4=5x^2\]\[x^4-5x^2+4=0\]

Answer 2

use substitution\[x^2=t\]
Then solve
\[t^2-5t+4=0\]

Answer 3

(x^2-4)(x^2-1) = 0

Answer 4

\[(x^2-4)(x^2-1)=0\]\[x^2=4\]\[x^2=1\]

Answer 5

i.e x = 2,-2,1,-1

Answer 6

Its a bi quadratic equation .Thus four possible roots...

i.e x = 2,-2,1,-1

Answer 7

there will be 4 possible roots: 2, -2 ,1,-1
Hope could help you !

Answer 8

Thanks this helped a lot (:

Answer 9

sub y as x power -2 and solve. x=2,-2,1,-1

Answer 10

let y= x ^-2, so eqn is 4y^2 + 1 = 5y
(4y-1)(y-1)=0, y=1, 1/4
x^-2 = 1 ,1/4, so x= 1,-1,2,-2