finding roots?

Question

2x^2 = (sqrt of 3x) -1/2
2x^2 = 3ix -1/2
2x^2 -3ix=-1/2
x(2x-3i)=1/2

and

4=8x^-2 +x^-4
4=x^-2(8+x^-2)

these two problems are giving me trouble and i dont know how to go on from there..like finding the roots,i appreciate any help thanks

Answer 1

2x^2 = (sqrt of 3x) -1/2
2x^2 +1/2 = sqrt(3x)
4x^4 +2x^2 +1/4 = 3x
4x^4+2x^2-3x+1/4 = 0
This has roots at x=.38 and x = 1.48 approximately

4=8x^-2 +x^-4
4=8/x^2 +1/x^4
4x^4 =8x^2 +1
4x^4 -8x^2 -1=0
Let u = x^2
Then 4u^2 -8u -1 = 0
Use quadratic formula to get u = 1 +/- .5sqrt(5)
So x^2 = 1+/- .5sqrt(5)
x = +/- sqrt[1 +/- .5sqrt(5)]
You should get 4 real solutions from the above

Answer 2

2x^2 = (sqrt of 3x) -1/2 won't give
2x^2 = 3ix -1/2

If it's
2x^2 = (sqrt of 3)x -1/2 then
2x^2 - (sqrt of 3)x +1/2=0, use quadratic formula

If it's 2x^2 = (sqrt of 3x) -1/2
then 2x^2 +1/2 = (sqrt of 3x) and square...
but then you get a 4th degree polynomial without rational roots

Multiplying 4=8/x^2 +1/x^4 by x^4 gives
4x^4=8x^2+1
4x^4-8x^2-1=0
4(x^2)^2-8x^2-1=0, use quadratic formula (y=x^2)

Answer 3

2x^2 = (sqrt of 3x) -1/2
2x^2 + 1/2= (sqrt of 3x).....sq both sides


and

4=8x^-2 +x^-4
=> 4 = 8/x^2 +1/x^4.........X by LCM
=> 4 x^4 = 8 x^2 + 1
=> 4 x^4 - 8 x^2 - 1 = 0