Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation 2x^2-x=0
-
Equation 4*x^2-13*x+3=0
-
Equation -x^2+6*x-7=0
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Equation x^3-2x+1=0
- Express {x} in terms of y where:
- -6*x-15*y=3
- -9*x+7*y=8
- -8*x-8*y=8
- -17*x-16*y=6
- Derivative of:
-
x^3-2x+1
- Graphing y =:
- x^3-2x+1
-
Identical expressions
- x^ three -2x+ one = zero
- x cubed minus 2x plus 1 equally 0
- x to the power of three minus 2x plus one equally zero
- x3-2x+1=0
- x³-2x+1=0
- x to the power of 3-2x+1=0
- x^3-2x+1=O
-
Similar expressions
- x^3-2x-1=0
- x^3+2x+1=0
x^3-2x+1=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$\left(x^{3} - 2 x\right) + 1 = 0$$
transform
$$\left(- 2 x + \left(x^{3} - 1\right)\right) + 2 = 0$$
or
$$\left(- 2 x + \left(x^{3} - 1^{3}\right)\right) + 2 = 0$$
$$- 2 \left(x - 1\right) + \left(x^{3} - 1^{3}\right) = 0$$
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) - 2 \left(x - 1\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(\left(x^{2} + x\right) + 1^{2}\right) - 2\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + x - 1\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + x - 1 = 0$$
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 two roots.
or
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}$$
The final answer for x^3 - 2*x + 1 = 0:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}$$
$$\left(x^{3} - 2 x\right) + 1 = 0$$
transform
$$\left(- 2 x + \left(x^{3} - 1\right)\right) + 2 = 0$$
or
$$\left(- 2 x + \left(x^{3} - 1^{3}\right)\right) + 2 = 0$$
$$- 2 \left(x - 1\right) + \left(x^{3} - 1^{3}\right) = 0$$
$$\left(x - 1\right) \left(\left(x^{2} + x\right) + 1^{2}\right) - 2 \left(x - 1\right) = 0$$
Take common factor -1 + x from the equation
we get:
$$\left(x - 1\right) \left(\left(\left(x^{2} + x\right) + 1^{2}\right) - 2\right) = 0$$
or
$$\left(x - 1\right) \left(x^{2} + x - 1\right) = 0$$
then:
$$x_{1} = 1$$
and also
we get the equation
$$x^{2} + x - 1 = 0$$
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) = 5
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}$$
The final answer for x^3 - 2*x + 1 = 0:
$$x_{1} = 1$$
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}$$
$$x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -2$$
$$v = \frac{d}{a}$$
$$v = 1$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -2$$
$$x_{1} x_{2} x_{3} = 1$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -2$$
$$v = \frac{d}{a}$$
$$v = 1$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -2$$
$$x_{1} x_{2} x_{3} = 1$$
Sum and product of roots
[src]
sum
___ ___
1 \/ 5 1 \/ 5
1 + - - + ----- + - - - -----
2 2 2 2
$$\left(- \frac{\sqrt{5}}{2} - \frac{1}{2}\right) + \left(\left(- \frac{1}{2} + \frac{\sqrt{5}}{2}\right) + 1\right)$$
=
0
$$0$$
product
/ ___\ / ___\ | 1 \/ 5 | | 1 \/ 5 | |- - + -----|*|- - - -----| \ 2 2 / \ 2 2 /
$$\left(- \frac{1}{2} + \frac{\sqrt{5}}{2}\right) \left(- \frac{\sqrt{5}}{2} - \frac{1}{2}\right)$$
=
-1
$$-1$$
-1
Rapid solution
[src]
x1 = 1
$$x_{1} = 1$$
___
1 \/ 5
x2 = - - + -----
2 2
$$x_{2} = - \frac{1}{2} + \frac{\sqrt{5}}{2}$$
___
1 \/ 5
x3 = - - - -----
2 2
$$x_{3} = - \frac{\sqrt{5}}{2} - \frac{1}{2}$$
x3 = -sqrt(5)/2 - 1/2
Numerical answer
[src]
x1 = 1.0
x2 = 0.618033988749895
x3 = -1.61803398874989
x3 = -1.61803398874989
The graph