x^4+4x^2-5=0?

Answer 1

x^2 = y
y^2 + 4y - 5 = 0
(y+5)(y-1) = 0
y = -5
y = 1

x = sqrt(-5) = ±sqrt(5)i
x = sqrt(1) = ±1

x = {sqrt(5)i, -sqrt(5)i, -1, 1}

Answer 2

x=1

Answer 3

x^4 + 4x^2 -5 =0
Let m = x^2.
m^2 +4m -5 =0

Using the quadratic formula,
x =[ -b (+-)√(b^2 - 4ac)]/ 2a

Substitute a= 1, b = 4, c =-5 into the above formula, we get
i.e. m ={ -4 (+-)√[(4)^2 - 4*1*(-5)]} / (2*1)
m = -4 (+-) √(16+20) / 2
m = [-4 (+-) √36] /2
m = [-4(+-) 6] / 2
m = (-4+6)/2 or (-4-6)/2
m = 1 or m = -5

Since m=x^ 2 =1 and m=x^2=-5,
so x^2 =1 and x^2 = -5 (rejected)#
x = (+-)√1
x = (+-) 1

# we reject x^2 = -5 solution because it is impossible to take the SQUARE root of a negative number.

Answer 4

(x² + 5).(x² - 1) = 0
(x ² + 5).(x - 1).(x + 1) = 0
x² + 5 = 0 , x = 1 , x = - 1
x² = - 5
x² = 5i²
x = (±√5).i , x = 1 , x = - 1