Mister Exam
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How to use it?
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Identical expressions
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Similar expressions
- z^4+2z^2+4=0
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z^4-2z^2+4=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(z^{4} - 2 z^{2}\right) + 4 = 0$$
Do replacement
$$v = z^{2}$$
then the equation will be the:
$$v^{2} - 2 v + 4 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 4$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$v_{1} = 1 + \sqrt{3} i$$
$$v_{2} = 1 - \sqrt{3} i$$
The final answer:
Because
$$v = z^{2}$$
then
$$z_{1} = \sqrt{v_{1}}$$
$$z_{2} = - \sqrt{v_{1}}$$
$$z_{3} = \sqrt{v_{2}}$$
$$z_{4} = - \sqrt{v_{2}}$$
then:
$$z_{1} = $$
$$\frac{0}{1} + \frac{\left(1 + \sqrt{3} i\right)^{\frac{1}{2}}}{1} = \sqrt{1 + \sqrt{3} i}$$
$$z_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(1 + \sqrt{3} i\right)^{\frac{1}{2}}}{1} = - \sqrt{1 + \sqrt{3} i}$$
$$z_{3} = $$
$$\frac{0}{1} + \frac{\left(1 - \sqrt{3} i\right)^{\frac{1}{2}}}{1} = \sqrt{1 - \sqrt{3} i}$$
$$z_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(1 - \sqrt{3} i\right)^{\frac{1}{2}}}{1} = - \sqrt{1 - \sqrt{3} i}$$
$$\left(z^{4} - 2 z^{2}\right) + 4 = 0$$
Do replacement
$$v = z^{2}$$
then the equation will be the:
$$v^{2} - 2 v + 4 = 0$$
This equation is of the form
a*v^2 + b*v + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$v_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$v_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2$$
$$c = 4$$
, then
D = b^2 - 4 * a * c =
(-2)^2 - 4 * (1) * (4) = -12
Because D<0, then the equation
has no real roots,
but complex roots is exists.
v1 = (-b + sqrt(D)) / (2*a)
v2 = (-b - sqrt(D)) / (2*a)
or
$$v_{1} = 1 + \sqrt{3} i$$
$$v_{2} = 1 - \sqrt{3} i$$
The final answer:
Because
$$v = z^{2}$$
then
$$z_{1} = \sqrt{v_{1}}$$
$$z_{2} = - \sqrt{v_{1}}$$
$$z_{3} = \sqrt{v_{2}}$$
$$z_{4} = - \sqrt{v_{2}}$$
then:
$$z_{1} = $$
$$\frac{0}{1} + \frac{\left(1 + \sqrt{3} i\right)^{\frac{1}{2}}}{1} = \sqrt{1 + \sqrt{3} i}$$
$$z_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(1 + \sqrt{3} i\right)^{\frac{1}{2}}}{1} = - \sqrt{1 + \sqrt{3} i}$$
$$z_{3} = $$
$$\frac{0}{1} + \frac{\left(1 - \sqrt{3} i\right)^{\frac{1}{2}}}{1} = \sqrt{1 - \sqrt{3} i}$$
$$z_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(1 - \sqrt{3} i\right)^{\frac{1}{2}}}{1} = - \sqrt{1 - \sqrt{3} i}$$
Rapid solution
[src]
___ ___
\/ 6 I*\/ 2
z1 = - ----- - -------
2 2
$$z_{1} = - \frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 6 I*\/ 2
z2 = - ----- + -------
2 2
$$z_{2} = - \frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}$$
___ ___
\/ 6 I*\/ 2
z3 = ----- - -------
2 2
$$z_{3} = \frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}$$
___ ___
\/ 6 I*\/ 2
z4 = ----- + -------
2 2
$$z_{4} = \frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}$$
z4 = sqrt(6)/2 + sqrt(2)*i/2
Sum and product of roots
[src]
sum
___ ___ ___ ___ ___ ___ ___ ___
\/ 6 I*\/ 2 \/ 6 I*\/ 2 \/ 6 I*\/ 2 \/ 6 I*\/ 2
- ----- - ------- + - ----- + ------- + ----- - ------- + ----- + -------
2 2 2 2 2 2 2 2
$$\left(\left(\frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(\left(- \frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}\right) + \left(- \frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}\right)\right)\right) + \left(\frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
0
$$0$$
product
/ ___ ___\ / ___ ___\ / ___ ___\ / ___ ___\ | \/ 6 I*\/ 2 | | \/ 6 I*\/ 2 | |\/ 6 I*\/ 2 | |\/ 6 I*\/ 2 | |- ----- - -------|*|- ----- + -------|*|----- - -------|*|----- + -------| \ 2 2 / \ 2 2 / \ 2 2 / \ 2 2 /
$$\left(- \frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}\right) \left(- \frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}\right) \left(\frac{\sqrt{6}}{2} - \frac{\sqrt{2} i}{2}\right) \left(\frac{\sqrt{6}}{2} + \frac{\sqrt{2} i}{2}\right)$$
=
4
$$4$$
4
Numerical answer
[src]
z1 = -1.22474487139159 + 0.707106781186548*i
z2 = -1.22474487139159 - 0.707106781186548*i
z3 = 1.22474487139159 + 0.707106781186548*i
z4 = 1.22474487139159 - 0.707106781186548*i
z4 = 1.22474487139159 - 0.707106781186548*i