Mister Exam
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How to use it?
- Equation:
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Equation z^3+6z+20=0
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Identical expressions
- z^ three +6z+ twenty = zero
- z cubed plus 6z plus 20 equally 0
- z to the power of three plus 6z plus twenty equally zero
- z3+6z+20=0
- z³+6z+20=0
- z to the power of 3+6z+20=0
- z^3+6z+20=O
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Similar expressions
- z^3-6z+20=0
- z^3+6z-20=0
z^3+6z+20=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$z^{3} + 6 z + 20 = 0$$
transform
$$z^{3} + 6 z + 20 = 0$$
or
$$z^{3} + 6 z + 20 = 0$$
$$z^{3} + 6 z + 20 = 0$$
$$\left(z + 2\right) \left(z^{2} - 2 z + 4\right) + 6 z + 12 = 0$$
Take common factor $z + 2$ from the equation
we get:
$$\left(z + 2\right) \left(z^{2} - 2 z + 10\right) = 0$$
or
$$\left(z + 2\right) \left(z^{2} - 2 z + 10\right) = 0$$
then:
$$z_{1} = -2$$
and also
we get the equation
$$z^{2} - 2 z + 10 = 0$$
This equation is of the form
$$a*z^2 + b*z + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$z_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 10$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 10 + \left(-2\right)^{2} = -36$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$z_2 = \frac{(-b + \sqrt{D})}{2 a}$$
$$z_3 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$z_{2} = 1 + 3 i$$
Simplify
$$z_{3} = 1 - 3 i$$
Simplify
The final answer for (z^3 + 6*z + 20) + 0 = 0:
$$z_{1} = -2$$
$$z_{2} = 1 + 3 i$$
$$z_{3} = 1 - 3 i$$
$$z^{3} + 6 z + 20 = 0$$
transform
$$z^{3} + 6 z + 20 = 0$$
or
$$z^{3} + 6 z + 20 = 0$$
$$z^{3} + 6 z + 20 = 0$$
$$\left(z + 2\right) \left(z^{2} - 2 z + 4\right) + 6 z + 12 = 0$$
Take common factor $z + 2$ from the equation
we get:
$$\left(z + 2\right) \left(z^{2} - 2 z + 10\right) = 0$$
or
$$\left(z + 2\right) \left(z^{2} - 2 z + 10\right) = 0$$
then:
$$z_{1} = -2$$
and also
we get the equation
$$z^{2} - 2 z + 10 = 0$$
This equation is of the form
$$a*z^2 + b*z + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$z_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 10$$
, then
$$D = b^2 - 4 * a * c = $$
$$\left(-1\right) 1 \cdot 4 \cdot 10 + \left(-2\right)^{2} = -36$$
Because D<0, then the equation
has no real roots,
but complex roots is exists.
$$z_2 = \frac{(-b + \sqrt{D})}{2 a}$$
$$z_3 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$z_{2} = 1 + 3 i$$
Simplify
$$z_{3} = 1 - 3 i$$
Simplify
The final answer for (z^3 + 6*z + 20) + 0 = 0:
$$z_{1} = -2$$
$$z_{2} = 1 + 3 i$$
$$z_{3} = 1 - 3 i$$
Vieta's Theorem
it is reduced cubic equation
$$p z^{2} + z^{3} + q z + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 6$$
$$v = \frac{d}{a}$$
$$v = 20$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 6$$
$$z_{1} z_{2} z_{3} = 20$$
$$p z^{2} + z^{3} + q z + v = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = 6$$
$$v = \frac{d}{a}$$
$$v = 20$$
Vieta Formulas
$$z_{1} + z_{2} + z_{3} = - p$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = q$$
$$z_{1} z_{2} z_{3} = v$$
$$z_{1} + z_{2} + z_{3} = 0$$
$$z_{1} z_{2} + z_{1} z_{3} + z_{2} z_{3} = 6$$
$$z_{1} z_{2} z_{3} = 20$$
Sum and product of roots
[src]
sum
-2 + 1 - 3*I + 1 + 3*I
$$\left(-2\right) + \left(1 - 3 i\right) + \left(1 + 3 i\right)$$
=
0
$$0$$
product
-2 * 1 - 3*I * 1 + 3*I
$$\left(-2\right) * \left(1 - 3 i\right) * \left(1 + 3 i\right)$$
=
-20
$$-20$$
Rapid solution
[src]
z_1 = -2
$$z_{1} = -2$$
z_2 = 1 - 3*I
$$z_{2} = 1 - 3 i$$
z_3 = 1 + 3*I
$$z_{3} = 1 + 3 i$$
The graph