Solve: x^4+x^2-6=0?

Answer 1

Let t = x^2

t^2 + t - 6 = 0
(t - 2)(t + 3) = 0
t = 2 or t = -3

x^2 = - 3 --> x = +-sqrt(3)i
x^2 = 2 --> x = +-sqrt(2)

Answer 2

x⁴+ x² - 6 = 0

(x²+3) (x²-2) = 0

x = ±i√3
x = ±√2

Answer 3

Put A = x^2, then
A^2 + A - 6 = 0
A' = [-1 + sqrt.(1 - 4(-6))]/2 = [-1 + sqrt.(25)]/2 = [-1 + 5]/2
A' = 2
A'' = [-1 - sqrt.(1 - 4(-6))]/2 = [-1 - sqrt.(25)]/2 = [-1 - 5]/2
A'' = -3
So,
x^2 = 2 ==> x = +sqrt.(2) and x = -sqrt.(2)
x^2 = -3 = 3*i^2 ==> x = i*sqrt.(3), x = -i*sqrt.(3)

Answer 4

similar as above solution, but slightly different:

x1/2=±sqrt(2)
x3/4=±sqrt(3)i

and it goes like this:

x^2=t
so
t^2 + t - 6 = 0

t1/2 = (-b ± sqrt(b^2 - 4*a*c))/2*a a = 1, b = 1, c = -6
t1/2 = (-1 ± sqrt(1+24))/2
t1/2 = (-1 ± sqrt(25))/2
t1/2 = (-1 ± 5)/2
t1 = (-1+5)/2
t1 = 2
t2 = (-1-5)/2
t2 = -3

x1/2^2 = 2
x1/2 = ±sqrt(2) = ±1.4142... ... ...

x3/4^2 = -3
x3/4 = ±sqrt(-3)
x3/4 = ±sqrt(3)i

voila!

Answer 5

Let u = x^2.
Solve:
u^2 + u - 6 = 0
(u + 3)(u - 2) = 0
Substitute x^2 for u.
(x^2 + 3)(x^2 - 2) = 0
x^2 = -3
x = (+/-) sq rt (-3) = (+/-)(sq rt 3) i
x^2 = 2
x = (+/-) sq rt (2).

Answer 6

(x^2 + 3)(x^2 - 2) = 0
So:
x^2 + 3=0
x^2 = -3
x = (+ or -)sqrt(-3)
x = (+ or -)sqrt[3*(-1)]
x = (+ or -)sqrt(3)*sqrt(-1)
Note: sqrt(-1) is written as "i"
x = (+ or -)[ i * sqrt(3)]


OR:
x^2 - 2 = 0
x^2 = 2
x = (+ or -)sqrt(2)

VOILA!