Mister Exam
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Identical expressions
- z^ two + six *z+ thirteen = zero
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Similar expressions
- z^2+6*z-13=0
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z^2+6*z+13=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
$$a = 1$$
$$b = 6$$
$$c = 13$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$z_{1} = -3 + 2 i$$
$$z_{2} = -3 - 2 i$$
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
$$a = 1$$
$$b = 6$$
$$c = 13$$
, then
D = b^2 - 4 * a * c =
(6)^2 - 4 * (1) * (13) = -16
Because D<0, then the equation
has no real roots,
but complex roots is exists.
z1 = (-b + sqrt(D)) / (2*a)
z2 = (-b - sqrt(D)) / (2*a)
or
$$z_{1} = -3 + 2 i$$
$$z_{2} = -3 - 2 i$$
Vieta's Theorem
it is reduced quadratic equation
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 6$$
$$q = \frac{c}{a}$$
$$q = 13$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = -6$$
$$z_{1} z_{2} = 13$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 6$$
$$q = \frac{c}{a}$$
$$q = 13$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = -6$$
$$z_{1} z_{2} = 13$$
Sum and product of roots
[src]
sum
-3 - 2*I + -3 + 2*I
$$\left(-3 - 2 i\right) + \left(-3 + 2 i\right)$$
=
-6
$$-6$$
product
(-3 - 2*I)*(-3 + 2*I)
$$\left(-3 - 2 i\right) \left(-3 + 2 i\right)$$
=
13
$$13$$
13
Rapid solution
[src]
z1 = -3 - 2*I
$$z_{1} = -3 - 2 i$$
z2 = -3 + 2*I
$$z_{2} = -3 + 2 i$$
z2 = -3 + 2*i
The graph