Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 2*x^2+6*x+9=0
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Equation -x-2+3(x-3)=3(4-x)-3
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Equation x^2+11=0
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Equation 3*x^2-12=0
- Express {x} in terms of y where:
- -7*x-7*y=-14
- -6*x-20*y=8
- 10*x-2*y=0
- -17*x+2*y=9
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Identical expressions
- z^ four + ten *z^ two + one hundred and sixty-nine = zero
- z to the power of 4 plus 10 multiply by z squared plus 169 equally 0
- z to the power of four plus ten multiply by z to the power of two plus one hundred and sixty minus nine equally zero
- z4+10*z2+169=0
- z⁴+10*z²+169=0
- z to the power of 4+10*z to the power of 2+169=0
- z^4+10z^2+169=0
- z4+10z2+169=0
- z^4+10*z^2+169=O
-
Similar expressions
- z^4-10*z^2+169=0
- z^4+10*z^2-169=0
z^4+10*z^2+169=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(z^{4} + 10 z^{2}\right) + 169 = 0$$
Do replacement
$$v = z^{2}$$
then the equation will be the:
$$v^{2} + 10 v + 169 = 0$$
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
$$a = 1$$
$$b = 10$$
$$c = 169$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$v_{1} = -5 + 12 i$$
$$v_{2} = -5 - 12 i$$
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:
$$z_{1} = $$
$$\frac{0}{1} + \frac{\left(-5 + 12 i\right)^{\frac{1}{2}}}{1} = 2 + 3 i$$
$$z_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-5 + 12 i\right)^{\frac{1}{2}}}{1} = -2 - 3 i$$
$$z_{3} = $$
$$\frac{0}{1} + \frac{\left(-5 - 12 i\right)^{\frac{1}{2}}}{1} = 2 - 3 i$$
$$z_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-5 - 12 i\right)^{\frac{1}{2}}}{1} = -2 + 3 i$$
$$\left(z^{4} + 10 z^{2}\right) + 169 = 0$$
Do replacement
$$v = z^{2}$$
then the equation will be the:
$$v^{2} + 10 v + 169 = 0$$
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
$$a = 1$$
$$b = 10$$
$$c = 169$$
, then
D = b^2 - 4 * a * c =
(10)^2 - 4 * (1) * (169) = -576
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
$$v_{1} = -5 + 12 i$$
$$v_{2} = -5 - 12 i$$
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:
$$z_{1} = $$
$$\frac{0}{1} + \frac{\left(-5 + 12 i\right)^{\frac{1}{2}}}{1} = 2 + 3 i$$
$$z_{2} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-5 + 12 i\right)^{\frac{1}{2}}}{1} = -2 - 3 i$$
$$z_{3} = $$
$$\frac{0}{1} + \frac{\left(-5 - 12 i\right)^{\frac{1}{2}}}{1} = 2 - 3 i$$
$$z_{4} = $$
$$\frac{0}{1} + \frac{\left(-1\right) \left(-5 - 12 i\right)^{\frac{1}{2}}}{1} = -2 + 3 i$$
Sum and product of roots
[src]
sum
-2 - 3*I + -2 + 3*I + 2 - 3*I + 2 + 3*I
$$\left(\left(2 - 3 i\right) + \left(\left(-2 - 3 i\right) + \left(-2 + 3 i\right)\right)\right) + \left(2 + 3 i\right)$$
=
0
$$0$$
product
(-2 - 3*I)*(-2 + 3*I)*(2 - 3*I)*(2 + 3*I)
$$\left(-2 - 3 i\right) \left(-2 + 3 i\right) \left(2 - 3 i\right) \left(2 + 3 i\right)$$
=
169
$$169$$
169
Rapid solution
[src]
z1 = -2 - 3*I
$$z_{1} = -2 - 3 i$$
z2 = -2 + 3*I
$$z_{2} = -2 + 3 i$$
z3 = 2 - 3*I
$$z_{3} = 2 - 3 i$$
z4 = 2 + 3*I
$$z_{4} = 2 + 3 i$$
z4 = 2 + 3*i
Numerical answer
[src]
z1 = -2.0 - 3.0*i
z2 = 2.0 - 3.0*i
z3 = -2.0 + 3.0*i
z4 = 2.0 + 3.0*i
z4 = 2.0 + 3.0*i