Mister Exam
Other calculators
-
How to use it?
- Equation:
-
Equation x+(x*3)=168
-
Equation t-4/25=8
-
Equation x^2+12*x=0
-
Equation x^3=4x
- Express {x} in terms of y where:
- 15*x+2*y=8
- 1*x+4*y=-11
- 5*x-18*y=-17
- 17*x-5*y=15
- Derivative of:
-
x^3
-
4x
- Integral of d{x}:
-
x^3
-
4x
- x^3
- Sum of series:
- 4x
-
Identical expressions
- x^ three =4x
- x cubed equally 4x
- x to the power of three equally 4x
- x3=4x
- x³=4x
- x to the power of 3=4x
-
Similar expressions
- (x+3)^4-4(x+3)^2-5=0
- x^2+14*x+24=0
x^3=4x equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{3} = 4 x$$
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:
$$x_{1} = 0$$
$$x_{2} = 2$$
$$x_{3} = -2$$
$$x^{3} = 4 x$$
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