solve the following equation?

Question

3x^4=2x^2+8

Answer 1

3x^4-2x^2-8=0
Quadratic formula with x^2 instead of x

a=3, b=-2, c=-8
[2+-sqrt(4-4(3)(-8)]/6
(2+-10)/6
x^2=2 or x^2=-4/3

x=+2 or -2
x can also equal +sqrt(-4/3) or -sqrt(-4/3) but these are complex numbers

Answer 2

3x^4=2x^2+8 subtract 2x^2 + 8 from each side
3x^4-2x^2-8=0
(3x^2+4)(x^2-2)=0
(3x+4)(x+√2)(x-√2)
x=√2
x=-√2 are the only real roots

x=1/2 √3 i
x=-1/2 √3 i
are the imaginary roots.

Answer 3

The equation has 4 roots:
(3x^2 + 4)(x^2 - 2) = 0
(3x^2 + 4)(x + 2)(x - 2) = 0
x = -√2, +√2, -i2/√3, i2/√3, where i = √-1

Answer 4

3x^4 = 2x^2 + 8
3x^4 - 2x^2 - 8 = 0
(3x^2 + 4)(x^2 - 2) = 0

3x^2 + 4 = 0
3x^2 = - 4
x^2 = - 4/3
x = +- 2sqrt(i/3)

x^2 - 2 = 0
x^2 = 2
x = +- sqrt(2)

Check

3(2sqrt(i/3))^4 = 2(2sqrt(i/3))^2 + 8
3(16)/9 = 2(-4/3) + 8
16/3 = - 8/3 + 24/3
16/3 = 16/3

3(sqrt(2))^4 = 2(sqrt(2))^2 + 8
3(4) = 2(2) + 8
12 = 4 + 8
12 = 12

Answer 5

3x^4-2x^2-8=0
put x^2=t
the equation will now be
3t^2-2t-8=0
(3t+4)(t-2)=0
t=2 or -4/3
x^2=2
x=rt2
negative value inadmisible

Answer 6

taking x^2 = y
the eqn is now:
3y^2 = 2y + 8
=> 3y^2 - 2y - 8 = 0
=> 3y^2 -6y +4y -8 = 0
=> 3y(y - 2) +4(y - 2) = 0
=> y = 2 or y = -4/3
subs x^2 = y
=> x = + - 2^0.5 and + - 2/(3^.5)*i, where i = (-1)^.5

Answer 7

wat math are you in?