Mister Exam
Other calculators
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How to use it?
- Equation:
-
Equation x^2+2x+1=0
-
Equation 3x^2=2x+x^3
-
Equation cos(x)+1=0
- Equation 3*x+11/9=x-91/27
- Express {x} in terms of y where:
- -15*x+18*y=-6
- -2*x+1*y=-3
- -18*x-18*y=-16
- -16*x-7*y=-20
- Graphing y =:
-
3x^2
- 2x+x^3
- Derivative of:
-
3x^2
- Canonical form:
- 3x^2
-
Identical expressions
- three x^ two =2x+x^3
- 3x squared equally 2x plus x cubed
- three x to the power of two equally 2x plus x cubed
- 3x2=2x+x3
- 3x²=2x+x³
- 3x to the power of 2=2x+x to the power of 3
-
Similar expressions
- 3x^2=2x-x^3
3x^2=2x+x^3 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Given the equation:
$$3 x^{2} = x^{3} + 2 x$$
transform
Take common factor x from the equation
we get:
$$x \left(- x^{2} + 3 x - 2\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$- x^{2} + 3 x - 2 = 0$$
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
$$a = -1$$
$$b = 3$$
$$c = -2$$
, then
Because D > 0, then the equation has two roots.
or
$$x_{2} = 1$$
$$x_{3} = 2$$
The final answer for 3*x^2 - 2*x - x^3 = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = 2$$
$$3 x^{2} = x^{3} + 2 x$$
transform
Take common factor x from the equation
we get:
$$x \left(- x^{2} + 3 x - 2\right) = 0$$
then:
$$x_{1} = 0$$
and also
we get the equation
$$- x^{2} + 3 x - 2 = 0$$
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
$$a = -1$$
$$b = 3$$
$$c = -2$$
, then
D = b^2 - 4 * a * c =
(3)^2 - 4 * (-1) * (-2) = 1
Because D > 0, then the equation has two roots.
x2 = (-b + sqrt(D)) / (2*a)
x3 = (-b - sqrt(D)) / (2*a)
or
$$x_{2} = 1$$
$$x_{3} = 2$$
The final answer for 3*x^2 - 2*x - x^3 = 0:
$$x_{1} = 0$$
$$x_{2} = 1$$
$$x_{3} = 2$$
Vieta's Theorem
rewrite the equation
$$3 x^{2} = x^{3} + 2 x$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 3 x^{2} + 2 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -3$$
$$q = \frac{c}{a}$$
$$q = 2$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 3$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 2$$
$$x_{1} x_{2} x_{3} = 0$$
$$3 x^{2} = x^{3} + 2 x$$
of
$$a x^{3} + b x^{2} + c x + d = 0$$
as reduced cubic equation
$$x^{3} + \frac{b x^{2}}{a} + \frac{c x}{a} + \frac{d}{a} = 0$$
$$x^{3} - 3 x^{2} + 2 x = 0$$
$$p x^{2} + q x + v + x^{3} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -3$$
$$q = \frac{c}{a}$$
$$q = 2$$
$$v = \frac{d}{a}$$
$$v = 0$$
Vieta Formulas
$$x_{1} + x_{2} + x_{3} = - p$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = q$$
$$x_{1} x_{2} x_{3} = v$$
$$x_{1} + x_{2} + x_{3} = 3$$
$$x_{1} x_{2} + x_{1} x_{3} + x_{2} x_{3} = 2$$
$$x_{1} x_{2} x_{3} = 0$$
The graph