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k^2+1=0 equation

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Numerical solution:

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The solution

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 2        
k  + 1 = 0
k2+1=0k^{2} + 1 = 0
Simplify
Detail solution
This equation is of the form
a*k^2 + b*k + c = 0

A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
k1=D−b2ak_{1} = \frac{\sqrt{D} - b}{2 a}
k2=−D−b2ak_{2} = \frac{- \sqrt{D} - b}{2 a}
where D = b^2 - 4*a*c - it is the discriminant.
Because
a=1a = 1
b=0b = 0
c=1c = 1
, then
D = b^2 - 4 * a * c = 

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

Because D<0, then the equation
has no real roots,
but complex roots is exists.
k1 = (-b + sqrt(D)) / (2*a)

k2 = (-b - sqrt(D)) / (2*a)

or
k1=ik_{1} = i
Simplify
k2=−ik_{2} = - i
Simplify
Vieta's Theorem
it is reduced quadratic equation
k2+kp+q=0k^{2} + k p + q = 0
where
p=bap = \frac{b}{a}
p=0p = 0
q=caq = \frac{c}{a}
q=1q = 1
Vieta Formulas
k1+k2=−pk_{1} + k_{2} = - p
k1k2=qk_{1} k_{2} = q
k1+k2=0k_{1} + k_{2} = 0
k1k2=1k_{1} k_{2} = 1
Rapid solution [src] 
k1 = -I
k1=−ik_{1} = - i
Simplify 
k2 = I
k2=ik_{2} = i
Simplify
Sum and product of roots [src] 
sum
0 - I + I
(0−i)+i\left(0 - i\right) + i 
=
0
00 
product
1*-I*I
i1(−i)i 1 \left(- i\right) 
=
1
11 
1
Numerical answer [src] 
k1 = -1.0*i
k2 = 1.0*i
k2 = 1.0*i
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