Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation -x^2+5*x-6=0
-
Equation 1/(x-1)^2+2/(x-1)-3=0
-
Equation tan(x)=4
- Equation z^2+3+4*i=0
- Express {x} in terms of y where:
- 10*x-15*y=16
- -18*x-6*y=-1
- -1*x-17*y=-12
- -2*x+20*y=-2
-
Identical expressions
- z^ two + three + four *i= zero
- z squared plus 3 plus 4 multiply by i equally 0
- z to the power of two plus three plus four multiply by i equally zero
- z2+3+4*i=0
- z²+3+4*i=0
- z to the power of 2+3+4*i=0
- z^2+3+4i=0
- z2+3+4i=0
- z^2+3+4*i=O
-
Similar expressions
- z^2+3-4*i=0
- z^2-3+4*i=0
z^2+3+4*i=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 3 + 4 i$$
, then
The equation has two roots.
or
$$z_{1} = 1 - 2 i$$
$$z_{2} = -1 + 2 i$$
a*z^2 + b*z + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 0$$
$$c = 3 + 4 i$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (3 + 4*i) = -12 - 16*i
The equation has two roots.
z1 = (-b + sqrt(D)) / (2*a)
z2 = (-b - sqrt(D)) / (2*a)
or
$$z_{1} = 1 - 2 i$$
$$z_{2} = -1 + 2 i$$
Vieta's Theorem
it is reduced quadratic equation
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 3 + 4 i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 0$$
$$z_{1} z_{2} = 3 + 4 i$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 3 + 4 i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 0$$
$$z_{1} z_{2} = 3 + 4 i$$
The graph
Sum and product of roots
[src]
sum
-1 + 2*I + 1 - 2*I
$$\left(1 - 2 i\right) + \left(-1 + 2 i\right)$$
=
0
$$0$$
product
(-1 + 2*I)*(1 - 2*I)
$$\left(-1 + 2 i\right) \left(1 - 2 i\right)$$
=
3 + 4*I
$$3 + 4 i$$
3 + 4*i
Rapid solution
[src]
z1 = -1 + 2*I
$$z_{1} = -1 + 2 i$$
z2 = 1 - 2*I
$$z_{2} = 1 - 2 i$$
z2 = 1 - 2*i