Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation 1+sin(x)=0
-
Equation tg(x)=1
-
Equation x^4+4=0
-
Equation -x-4+5(x+3)=5(-1-x)-2
- Express {x} in terms of y where:
- 4*x+2*y=5
- 1*x+10*y=-18
- -18*x-2*y=9
- -11*x+9*y=-20
-
Identical expressions
- 5x^ three -5x= zero
- 5x cubed minus 5x equally 0
- 5x to the power of three minus 5x equally zero
- 5x3-5x=0
- 5x³-5x=0
- 5x to the power of 3-5x=0
- 5x^3-5x=O
-
Similar expressions
- 5x^3+5x=0
5x^3-5x=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$5 x^{3} - 5 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(5 x^{2} - 5\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$5 x^{2} - 5 = 0$$
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
$$a = 5$$
$$b = 0$$
$$c = -5$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 1$$
Simplify
$$x_{3} = -1$$
Simplify
The final answer for (5*x^3 - 5*x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = -1$$
$$5 x^{3} - 5 x = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(5 x^{2} - 5\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$5 x^{2} - 5 = 0$$
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
$$a = 5$$
$$b = 0$$
$$c = -5$$
, then
D = b^2 - 4 * a * c =
(0)^2 - 4 * (5) * (-5) = 100
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 1$$
Simplify
$$x_{3} = -1$$
Simplify
The final answer for (5*x^3 - 5*x) + 0 = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = -1$$
Vieta's Theorem
rewrite the equation
$$5 x^{3} - 5 x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -1$$
$$x_{1} x_{2} x_{3} = 0$$
$$5 x^{3} - 5 x = 0$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 0$$
$$q = \frac{c}{a}$$
$$q = -1$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 0$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = -1$$
$$x_{1} x_{2} x_{3} = 0$$
Sum and product of roots
[src]
sum
0 - 1 + 0 + 1
$$\left(\left(-1 + 0\right) + 0\right) + 1$$
=
0
$$0$$
product
1*-1*0*1
$$1 \left(-1\right) 0 \cdot 1$$
=
0
$$0$$
0
The graph