Mister Exam
Other calculators
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How to use it?
- Equation:
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Equation x^2+4*x+4=0
-
Equation 90+x-4*10=60
-
Equation 6*3+x=72
-
Equation x^3+x^2=0
- Express {x} in terms of y where:
- -4*x+19*y=-13
- -5*x-6*y=-1
- -16*x-2*y=4
- -7*x+15*y=18
- Derivative of:
-
x^3+x^2
- Graphing y =:
- x^3+x^2
- Factor polynomial:
- x^3+x^2
-
Identical expressions
- x^ three +x^ two = zero
- x cubed plus x squared equally 0
- x to the power of three plus x to the power of two equally zero
- x3+x2=0
- x³+x²=0
- x to the power of 3+x to the power of 2=0
- x^3+x^2=O
-
Similar expressions
- x^3-x^2=0
x^3+x^2=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$x^{3} + x^{2} = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} + x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} + x = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = 0$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 0$$
$$x_{3} = -1$$
The final answer for x^3 + x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = -1$$
$$x^{3} + x^{2} = 0$$
transform
Take common factor x from the equation
we get:
$$x \left(x^{2} + x\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$x^{2} + x = 0$$
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
$$a = 1$$
$$b = 1$$
$$c = 0$$
, then
D = b^2 - 4 * a * c =
(1)^2 - 4 * (1) * (0) = 1
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 0$$
$$x_{3} = -1$$
The final answer for x^3 + x^2 = 0:
$$x_{1} = 0$$
$$x_{2} = 0$$
$$x_{3} = -1$$
Vieta's Theorem
it is reduced cubic equation
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 1$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -1$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = 1$$
$$q = \frac{c}{a}$$
$$q = 0$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = -1$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 0$$
$$x_{1} x_{2} x_{3} = 0$$
The graph