Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation 2+3x=-2x-13
-
Equation cos(x)=-(1/2)
-
Equation |x^2-1|-|x-3|=7
-
Equation 2^2-x-1=x^2-5*x-(-1-x^2)
- Express {x} in terms of y where:
- -1*x-7*y=5
- -6*x+2*y=-10
- 10*x-18*y=6
- 4*x-18*y=-14
- Graphing y =:
- x^3-4*x
- Integral of d{x}:
- x^3-4*x
- Factor polynomial:
- x^3-4*x
-
Identical expressions
- x^ three - four *x= zero
- x cubed minus 4 multiply by x equally 0
- x to the power of three minus four multiply by x equally zero
- x3-4*x=0
- x³-4*x=0
- x to the power of 3-4*x=0
- x^3-4x=0
- x3-4x=0
- x^3-4*x=O
-
Similar expressions
- x^3+4*x=0
x^3-4*x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{3} - 4 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 4\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 4 = 0$$
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
$$a = 1$$
$$b = 0$$
$$c = -4$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 2$$
$$x_{3} = -2$$
The final answer for x^3 - 4*x = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$x_{3} = -2$$
$$x^{3} - 4 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} - 4\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} - 4 = 0$$
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
$$a = 1$$
$$b = 0$$
$$c = -4$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (1) * (-4) = 16
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 2$$
$$x_{3} = -2$$
The final answer for x^3 - 4*x = 0:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$x_{3} = -2$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -4$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -4$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
-2 + 2
$$-2 + 2$$
=
0
$$0$$
product
-2*0*2
$$2 \left(- 0\right)$$
=
0
$$0$$
0
The graph