Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 9x-26=30-5x
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Equation x+19=12
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Equation 4*(x-3)=x+6
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Equation -x-7=x
- Express {x} in terms of y where:
- 17*x+20*y=-12
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- 10*x-18*y=-11
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Identical expressions
- 9x^ four - one = zero
- 9x to the power of 4 minus 1 equally 0
- 9x to the power of four minus one equally zero
- 9x4-1=0
- 9x⁴-1=0
- 9x^4-1=O
-
Similar expressions
- 9x^4+1=0
9x^4-1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation
$$9 x^{4} - 1 = 0$$
Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then
the equation has two real roots.
Get the root 4-th degree of the equation sides:
We get:
$$\sqrt[4]{9} \sqrt[4]{\left(1 x + 0\right)^{4}} = 1$$
$$\sqrt[4]{9} \sqrt[4]{\left(1 x + 0\right)^{4}} = -1$$
or
$$\sqrt{3} x = 1$$
$$\sqrt{3} x = -1$$
Expand brackets in the left part
Divide both parts of the equation by sqrt(3)
We get the answer: x = sqrt(3)/3
Expand brackets in the left part
Divide both parts of the equation by sqrt(3)
We get the answer: x = -sqrt(3)/3
or
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{4} = \frac{1}{9}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = \frac{1}{9}$$
where
$$r = \frac{\sqrt{3}}{3}$$
- the magnitude of the complex number
Substitute r:
$$e^{4 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = 1$$
so
$$\cos{\left(4 p \right)} = 1$$
and
$$\sin{\left(4 p \right)} = 0$$
then
$$p = \frac{\pi N}{2}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = - \frac{\sqrt{3}}{3}$$
$$z_{2} = \frac{\sqrt{3}}{3}$$
$$z_{3} = - \frac{\sqrt{3} i}{3}$$
$$z_{4} = \frac{\sqrt{3} i}{3}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
$$x_{3} = - \frac{\sqrt{3} i}{3}$$
$$x_{4} = \frac{\sqrt{3} i}{3}$$
$$9 x^{4} - 1 = 0$$
Because equation degree is equal to = 4 - contains the even number 4 in the numerator, then
the equation has two real roots.
Get the root 4-th degree of the equation sides:
We get:
$$\sqrt[4]{9} \sqrt[4]{\left(1 x + 0\right)^{4}} = 1$$
$$\sqrt[4]{9} \sqrt[4]{\left(1 x + 0\right)^{4}} = -1$$
or
$$\sqrt{3} x = 1$$
$$\sqrt{3} x = -1$$
Expand brackets in the left part
x*sqrt3 = 1
Divide both parts of the equation by sqrt(3)
x = 1 / (sqrt(3))
We get the answer: x = sqrt(3)/3
Expand brackets in the left part
x*sqrt3 = -1
Divide both parts of the equation by sqrt(3)
x = -1 / (sqrt(3))
We get the answer: x = -sqrt(3)/3
or
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
All other 2 root(s) is the complex numbers.
do replacement:
$$z = x$$
then the equation will be the:
$$z^{4} = \frac{1}{9}$$
Any complex number can presented so:
$$z = r e^{i p}$$
substitute to the equation
$$r^{4} e^{4 i p} = \frac{1}{9}$$
where
$$r = \frac{\sqrt{3}}{3}$$
- the magnitude of the complex number
Substitute r:
$$e^{4 i p} = 1$$
Using Euler’s formula, we find roots for p
$$i \sin{\left(4 p \right)} + \cos{\left(4 p \right)} = 1$$
so
$$\cos{\left(4 p \right)} = 1$$
and
$$\sin{\left(4 p \right)} = 0$$
then
$$p = \frac{\pi N}{2}$$
where N=0,1,2,3,...
Looping through the values of N and substituting p into the formula for z
Consequently, the solution will be for z:
$$z_{1} = - \frac{\sqrt{3}}{3}$$
$$z_{2} = \frac{\sqrt{3}}{3}$$
$$z_{3} = - \frac{\sqrt{3} i}{3}$$
$$z_{4} = \frac{\sqrt{3} i}{3}$$
do backward replacement
$$z = x$$
$$x = z$$
The final answer:
$$x_{1} = - \frac{\sqrt{3}}{3}$$
$$x_{2} = \frac{\sqrt{3}}{3}$$
$$x_{3} = - \frac{\sqrt{3} i}{3}$$
$$x_{4} = \frac{\sqrt{3} i}{3}$$
Sum and product of roots
[src]
sum
___ ___ ___ ___ -\/ 3 \/ 3 -I*\/ 3 I*\/ 3 ------- + ----- + --------- + ------- 3 3 3 3
$$\left(- \frac{\sqrt{3}}{3}\right) + \left(\frac{\sqrt{3}}{3}\right) + \left(- \frac{\sqrt{3} i}{3}\right) + \left(\frac{\sqrt{3} i}{3}\right)$$
=
0
$$0$$
product
___ ___ ___ ___ -\/ 3 \/ 3 -I*\/ 3 I*\/ 3 ------- * ----- * --------- * ------- 3 3 3 3
$$\left(- \frac{\sqrt{3}}{3}\right) * \left(\frac{\sqrt{3}}{3}\right) * \left(- \frac{\sqrt{3} i}{3}\right) * \left(\frac{\sqrt{3} i}{3}\right)$$
=
-1/9
$$- \frac{1}{9}$$
Rapid solution
[src]
___
-\/ 3
x_1 = -------
3
$$x_{1} = - \frac{\sqrt{3}}{3}$$
___
\/ 3
x_2 = -----
3
$$x_{2} = \frac{\sqrt{3}}{3}$$
___
-I*\/ 3
x_3 = ---------
3
$$x_{3} = - \frac{\sqrt{3} i}{3}$$
___
I*\/ 3
x_4 = -------
3
$$x_{4} = \frac{\sqrt{3} i}{3}$$
Numerical answer
[src]
x1 = -0.577350269189626*i
x2 = 0.577350269189626
x3 = -0.577350269189626
x4 = 0.577350269189626*i
x4 = 0.577350269189626*i
The graph