Mister Exam
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How to use it?
- Equation:
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Equation 13/(x+6)=2
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Equation x^2=18
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Equation -3(x-1)=24-10x
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Equation 15*x^2-4*x-3=0
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- Derivative of:
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(x-1)(x^2+x+1)
-
Identical expressions
- (x- one)(x^ two +x+ one)= zero
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- (x minus one)(x to the power of two plus x plus one) equally zero
- (x-1)(x2+x+1)=0
- x-1x2+x+1=0
- (x-1)(x²+x+1)=0
- (x-1)(x to the power of 2+x+1)=0
- x-1x^2+x+1=0
- (x-1)(x^2+x+1)=O
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Similar expressions
- (x+1)(x^2+x+1)=0
- (x-1)(x^2-x+1)=0
- (x-1)(x^2+x-1)=0
(x-1)(x^2+x+1)=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
You have entered
[src]
/ 2 \ (x - 1)*\x + x + 1/ = 0
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1\right) = 0$$
Detail solution
Given the equation:
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1\right) = 0$$
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 1 = 0$$
$$x^{2} + x + 1 = 0$$
solve the resulting equation:
1.
$$x - 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 1$$
We get the answer: x1 = 1
2.
$$x^{2} + x + 1 = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 1$$
$$c = 1$$
, then
Because D<0, then the equation
has no real roots,
but complex roots is exists.
or
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$x_{3} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
The final answer:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$x_{3} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1\right) = 0$$
Because the right side of the equation is zero, then the solution of the equation is exists if at least one of the multipliers in the left side of the equation equal to zero.
We get the equations
$$x - 1 = 0$$
$$x^{2} + x + 1 = 0$$
solve the resulting equation:
1.
$$x - 1 = 0$$
Move free summands (without x)
from left part to right part, we given:
$$x = 1$$
We get the answer: x1 = 1
2.
$$x^{2} + x + 1 = 0$$
This equation is of the form
a*x^2 + b*x + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$x_{2} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{3} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = 1$$
$$c = 1$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (1) = -3
Because D<0, then the equation
has no real roots,
but complex roots is exists.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$x_{3} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
The final answer:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
$$x_{3} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
Rapid solution
[src]
x1 = 1
$$x_{1} = 1$$
___
1 I*\/ 3
x2 = - - - -------
2 2
$$x_{2} = - \frac{1}{2} - \frac{\sqrt{3} i}{2}$$
___
1 I*\/ 3
x3 = - - + -------
2 2
$$x_{3} = - \frac{1}{2} + \frac{\sqrt{3} i}{2}$$
x3 = -1/2 + sqrt(3)*i/2
Sum and product of roots
[src]
sum
___ ___
1 I*\/ 3 1 I*\/ 3
1 + - - - ------- + - - + -------
2 2 2 2
$$\left(1 + \left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right)\right) + \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
0
$$0$$
product
/ ___\ / ___\ | 1 I*\/ 3 | | 1 I*\/ 3 | |- - - -------|*|- - + -------| \ 2 2 / \ 2 2 /
$$\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right)$$
=
1
$$1$$
1
Numerical answer
[src]
x1 = 1.0
x2 = -0.5 - 0.866025403784439*i
x3 = -0.5 + 0.866025403784439*i
x3 = -0.5 + 0.866025403784439*i
The graph